MATH 1071 — Calculus for Business & Economics

Example The demand equation for a 20 oz soda is \(p = 3-2x\), where \(p\) is the price and \(x\) is the number sold (in hundreds) in one week. Find the elasticity of demand at \(p = 1.25,\ 1.50,\ 2\). Is demand elastic, inelastic or unit elastic in each case? \[ p = 3-2x \;\Longrightarrow\; p - 3 = -2x \;\Longrightarrow\; x = -\tfrac12 p + \tfrac32. \] \[ \frac{dx}{dp} = -\frac12. \] \[ E(p) = -\left(-\frac12\right)\cdot\frac{p}{-\frac12p+\frac32} = \frac{p}{2\left(-\frac12p+\frac32\right)} = \frac{p}{3-p}. \]

Now evaluate:

\[ E(1.25) = \frac{1.25}{1.75} = \frac57 \approx 0.714, \qquad E(1.5) = \frac{1.5}{1.5} = 1, \qquad E(2) = \frac{2}{1} = 2. \]

So demand is inelastic at \(p = 1.25\), unit elastic at \(p = 1.50\), and elastic at \(p = 2\).

Example Suppose \(E(4) = \tfrac35\). What does this mean? Does increasing the price increase or decrease revenue?

\(E(4) = \dfrac{3\%\text{ decrease in demand}}{5\%\text{ increase in price}}\): when the price is $4, a 5% increase in price leads to (about) a 3% decrease in demand. Demand falls by a smaller percentage than price rises, so raising the price increases revenue.

Now suppose instead \(E(4) = 2 = \dfrac{2\%\text{ decrease in demand}}{1\%\text{ increase in price}}\). When the price is $4, a 1% increase in price leads to a 2% decrease in demand. We lose more customers than we gain in price, so raising the price decreases revenue.

Example (Maximizing Revenue) The demand for a product is \(x = (36-2p)^2\), where \(p\) is the price and \(x\) is the number of items sold, \(0\le p\le 18\). What should be charged to maximize revenue?

To maximize revenue, set \(E = 1\). By the chain rule,

\[ \frac{dx}{dp} = 2(36-2p)\cdot(-2) = -4(36-2p). \]

So

\[ E(p) = -\bigl(-4(36-2p)\bigr)\cdot\frac{p}{(36-2p)^2} = \frac{4p\,\cancel{(36-2p)}}{(36-2p)^{\cancel{2}}} = \frac{4p}{36-2p}. \]

Set \(E(p) = 1\):

\[ \frac{4p}{36-2p} = 1 \;\Longrightarrow\; 4p = 36 - 2p \;\Longrightarrow\; 6p = 36 \;\Longrightarrow\; p = 6. \]

Revenue is maximized at a price of $6. Check: graphing \(R(p) = p(36-2p)^2\) on \([0,18]\) shows a peak at \((6, 3456)\).

Example A small grocery store makes their own candy bars and finds \(E(p) = \dfrac{p}{5-p}\). How much should they charge to maximize revenue? \[ \frac{p}{5-p} = 1 \;\Longrightarrow\; p = 5 - p \;\Longrightarrow\; 2p = 5 \;\Longrightarrow\; p = 2.5. \]

Revenue is maximized when \(p = \)$2.50.

Example The price of season tickets is $40 and \(E(40) = 1.75\). If the price increases by $4, what change do we expect in demand? If demand was 1000, what would the new demand be?

Going from $40 to $44 is an increase of \(\frac{4}{40} = 0.10\), i.e. a 10% increase in price. Then

\[ \%\text{ decrease in demand} \approx 1.75 \times 10\% = 17.5\%. \]

New demand \(\approx 1000 - (0.175)(1000) = 1000 - 175 = 825\) tickets.

Example Demand is given by \(x(p) = 1200 - 6p^2\). Find \(x(10)\) and \(E(10)\). Use these to estimate \(x(11)\), then compare with the actual value. \[ x(10) = 1200 - 6(100) = 600. \] \[ \frac{dx}{dp} = -12p, \qquad E(p) = -(-12p)\cdot\frac{p}{1200-6p^2} = \frac{12p^2}{1200-6p^2}, \qquad E(10) = \frac{1200}{600} = 2. \]

When the price is $10, a 1% increase in price leads to about a 2% decrease in demand. Going from \(p = 10\) to \(p = 11\) is a \(\frac{1}{10} = 10\%\) increase in price, so we expect about a \(20\%\) decrease in demand:

\[ x(11) \approx 600 - (0.20)(600) = 600 - 120 = 480. \]

Check: the actual value is \(x(11) = 1200 - 6(121) = 474\). Pretty close!

Example (Warm-up) Let \(f(x) = \sqrt x\). Find the tangent line at \(x = 4\) and at \(x = 9\).

First, \(f'(x) = \tfrac12x^{-1/2} = \dfrac{1}{2\sqrt x}\).

• At \(x = 4\): point \((4, f(4)) = (4,2)\), slope \(f'(4) = \frac{1}{2\sqrt4} = \frac14\). Tangent line: \(y - 2 = \tfrac14(x-4)\), i.e. \(y = \tfrac14(x-4)+2\).
• At \(x = 9\): point \((9, f(9)) = (9,3)\), slope \(f'(9) = \frac{1}{2\sqrt9} = \frac16\). Tangent line: \(y - 3 = \tfrac16(x-9)\), i.e. \(y = \tfrac16(x-9)+3\).

Example Find the linear approximation of \(f(x) = (1+x)^4\) at \(x = 0\). Use it to approximate \((1.02)^4\), and compare with the actual value.

Point: \((0, f(0)) = (0,1)\). Slope: by the chain rule \(f'(x) = 4(1+x)^3\cdot 1\), so \(f'(0) = 4\).

\[ y - 1 = 4(x-0) \;\Longrightarrow\; L(x) = 4x+1. \]

Notice that \((1.02)^4 = f(0.02)\), and \(0.02\) is close to \(0\). So

\[ (1.02)^4 = f(0.02)\approx L(0.02) = 4(0.02)+1 = 1.08. \]

The actual value is \((1.02)^4 = 1.08243216\dots\), very close.

Example Let \(C(x) = x^2+4x+3\).

(a) Find the linearization of the cost function at \(x = 10\).

\[ C(10) = 100+40+3 = 143, \qquad C'(x) = 2x+4, \qquad C'(10) = 24. \] \[ y - 143 = 24(x-10) \;\Longrightarrow\; L(x) = 24(x-10)+143. \]

(b) Once we have made ten items, what is the approximate cost of the next item?

\(C'(10) = 24\): after making 10 items, the next item costs approximately $24. (The exact cost of the 11th item is \(C(11)-C(10) = 168 - 143 = \$25\).)

(c) Compare the linear approximation with the "cost of the next item" idea, for example by estimating the cost of 13 items.

The first ten items cost $143, and each of the next \(13-10 = 3\) items costs about $24:

\[ C(13)\approx 143 + 24(13-10) = 215 = L(13). \]

The two ideas give exactly the same computation. (Actual: \(C(13) = 224\).)

Example Assume \(C(20) = 53\) and \(C'(20) = 2\). Approximate \(C(22)\) two ways.

It costs $53 to make 20 items. The 21st costs about $2 and the 22nd costs about $2:

\[ C(22)\approx 53 + 2 + 2 = 53 + 2\cdot(22-20) = \$57. \]

Point \((20,53)\), slope \(2\): \(L(x) = 53 + 2(x-20)\), so \(L(22) = 53 + 2(2) = 57\). Same answer.

Example (Exam 3 Review) The cost in dollars to produce \(x\) tons of a product is \(C(x) = \sqrt{2x^2+25}\).

Marginal cost: by the chain rule,

\[ C'(x) = \frac{1}{2\sqrt{2x^2+25}}\cdot 4x = \frac{2x}{\sqrt{2x^2+25}}. \]

At \(x = 10\): \(C(10) = \sqrt{225} = 15\) and \(C'(10) = \frac{20}{15} = \frac43\). So after producing 10 tons, the next ton costs about \(\$1.33\).

Linear approximation at 10: \(L(x) = 15 + \tfrac43(x-10)\), so the cost to produce 11 tons is about \(L(11) = 15 + \tfrac43 \approx \$16.33\) — the cost of 10 tons plus the marginal cost of one more. (Actual: \(C(11) = \sqrt{267}\approx 16.34\).)

Example

• \(f(x) = x^2\) (on all real numbers) has a smallest value of \(0\) at \(x = 0\), but no largest value: the parabola keeps going up forever.
• \(f(x) = \dfrac{1}{x+1}\) on \([-3,-1)\cup(-1,2]\) has no largest or smallest value: near the vertical asymptote \(x=-1\) the outputs blow up to \(\infty\) on one side and \(-\infty\) on the other.

Example Find the critical points of \(f(x) = x^2-6x+14\). \[ f'(x) = 2x - 6 = 0 \;\Longrightarrow\; 2x = 6 \;\Longrightarrow\; x = 3. \]

The derivative is defined everywhere, so the only critical point is \(x = 3\).

Example Find the absolute extrema of \(f(x) = x^2-6x+14\) on \([2,5]\).

From before, the only critical point is \(x = 3\), which lies in \([2,5]\). Check it along with the endpoints \(x = 2\) and \(x = 5\), plugging into the original function:

\(x\)	\(f(x) = x^2-6x+14\)
2	\(4-12+14 = 6\)
3	\(9-18+14 = 5\)	min
5	\(25-30+14 = 9\)	max

The absolute max is 9 (at \(x = 5\)) and the absolute min is 5 (at \(x=3\)).

Example Find the absolute extrema of \(f(x) = \dfrac{x^3}{3}+\dfrac{x^2}{2}-6x+2\) on \([0,4]\). (The square brackets mean the interval is closed.) \[ f'(x) = \tfrac13\cdot 3x^2 + \tfrac12\cdot 2x - 6 = x^2+x-6 = (x-2)(x+3) = 0. \]

So \(x = 2\) or \(x = -3\). But \(-3\) is not between 0 and 4, so we throw it out.

\(x\)	\(f(x)\)
0 (endpoint)	\(2\)
2 (critical point)	\(\tfrac83 + 2 - 12 + 2 = -\tfrac{16}{3}\approx -5.33\)	min
4 (endpoint)	\(\tfrac{64}{3}+8-24+2 = \tfrac{22}{3}\approx 7.33\)	max

Max value of \(\tfrac{22}{3}\) at \(x = 4\); min value of \(-\tfrac{16}{3}\) at \(x = 2\).

Example Find the absolute extrema of \(f(x) = 4x^2-16x+5\) on \([1,4)\). \[ f'(x) = 8x - 16 = 0 \;\Longrightarrow\; x = 2. \]

\(x\)	\(f(x)\)
1	\(4-16+5 = -7\)
2	\(16-32+5 = -11\)	absolute min
4 (not included)	\(64-64+5 = 5\)	would be the max, but \(x=4\) is not included

The absolute min is \(-11\) at \(x=2\). There is no absolute max: the function gets closer and closer to 5 as \(x\to4\), but never reaches it.

Example What are the critical points of \(f(x) = e^x(x-3)^{2/3}\)?

By the product rule and the chain rule,

\[ f'(x) = e^x\cdot\tfrac23(x-3)^{-1/3}\cdot(1) + e^x(x-3)^{2/3} = e^x\left[\frac{2}{3(x-3)^{1/3}} + (x-3)^{2/3}\right]. \]

Where \(f'\) is undefined: the fraction has \((x-3)^{1/3}\) in the denominator, which is zero when \(x = 3\). Since \(f(3) = 0\) is defined, \(x = 3\) is a critical point.

Where \(f' = 0\): since \(e^x\) is never zero, the bracket must be zero:

\[ \begin{aligned} \frac{2}{3(x-3)^{1/3}} + (x-3)^{2/3} &= 0\\ \frac{2}{3(x-3)^{1/3}} &= -(x-3)^{2/3}\\ 2 &= -3(x-3)^{2/3}(x-3)^{1/3} \qquad\text{(multiply both sides by } 3(x-3)^{1/3})\\ 2 &= -3(x-3)\\ 2 &= -3x + 9 \;\Longrightarrow\; -7 = -3x \;\Longrightarrow\; x = \tfrac73. \end{aligned} \]

The critical points are \(x = 3\) and \(x = \tfrac73\).

Example Use the derivative of \(f(x) = x^3+3x^2-9x+3\) to help sketch its graph. \[ f'(x) = 3x^2 + 6x - 9 = 3(x^2+2x-3) = 3(x-1)(x+3) = 0. \]

The critical points are \(x = 1\) and \(x = -3\).

Interval	\((-\infty,-3)\)	\((-3,1)\)	\((1,\infty)\)
Test point	\(f'(-4) = 3(-5)(-1) = 15\)	\(f'(0) = -9\)	\(f'(2) = 3(1)(5) = 15\)
Sign of \(f'\)	\(+\)	\(-\)	\(+\)
\(f\)	increasing ↗	decreasing ↘	increasing ↗

Plugging the critical points into \(f\): \(f(-3) = -27+27+27+3 = 30\) and \(f(1) = 1+3-9+3 = -2\). So the graph rises to a peak at \((-3,30)\), falls to a valley at \((1,-2)\), then rises again. It crosses the \(y\)-axis at \(f(0) = 3\).

Example Let \(f(x) = -x^3 + \tfrac32x^2 + 18x\). Find the intervals where \(f\) is increasing and decreasing and the location of any local extrema. Then sketch a graph. \[ f'(x) = -3x^2 + 3x + 18 = -3(x^2 - x - 6) = -3(x+2)(x-3). \]

The critical points are \(x = -2\) and \(x = 3\).

Interval	\((-\infty,-2)\)	\((-2,3)\)	\((3,\infty)\)
Test point	\(f'(-3) = -3(-1)(-6) = -18\)	\(f'(0) = 18\)	\(f'(4) = -3(6)(1) = -18\)
Sign of \(f'\)	\(-\)	\(+\)	\(-\)
\(f\)	decreasing ↘	increasing ↗	decreasing ↘

• Increasing: \((-2,3)\). Decreasing: \((-\infty,-2)\cup(3,\infty)\).
• Local min at \(x = -2\) (\(-\) to \(+\)): \(f(-2) = 8 + 6 - 36 = -22\).
• Local max at \(x = 3\) (\(+\) to \(-\)): \(f(3) = -27 + 13.5 + 54 = 40.5\).

Sketch: coming down from the upper left, a valley at \((-2,-22)\), up through the origin to a peak at \((3, 40.5)\), then down to the lower right.

Example Let \(f(x) = 3x^2+6x+7\). Find where \(f\) is increasing and decreasing and the location of any local extrema. \[ f'(x) = 6x + 6 = 6(x+1) = 0 \;\Longrightarrow\; x = -1. \]

Test points: \(f'(-2) = -12+6 = -6 < 0\) and \(f'(0) = 6 > 0\). So \(f\) is decreasing on \((-\infty,-1)\) and increasing on \((-1,\infty)\), with a local min at \(x=-1\): \(f(-1) = 3-6+7 = 4\). (This is a parabola opening up with vertex \((-1,4)\), crossing the \(y\)-axis at 7.)

Example (Reading the Graph of \(f'\)) The graph of the derivative \(f'(x)\) is shown below. Find and classify all local extrema of \(f\).

We only care about the sign of \(f'\): above the \(x\)-axis means \(+\), below means \(-\). The graph of \(f'\) touches the \(x\)-axis at \(x = 2\), \(5\) and \(9\).

Interval	\((-\infty,2)\)	\((2,5)\)	\((5,9)\)	\((9,\infty)\)
Sign of \(f'\)	\(+\)	\(-\)	\(+\)	\(+\)
\(f\)	↗	↘	↗	↗

• \(x = 2\): \(+\) to \(-\), so a local max.
• \(x = 5\): \(-\) to \(+\), so a local min.
• \(x = 9\): \(f'(9) = 0\) but the sign stays \(+\), so no local min or max.

Example Determine the intervals of concavity for \(f(x) = x^3 + 6x^2 - 15x + 2\). \[ f'(x) = 3x^2+12x-15, \qquad f''(x) = 6x + 12 = 0 \;\Longrightarrow\; x = -2. \]

Interval	\((-\infty,-2)\)	\((-2,\infty)\)
Test point	\(f''(-3) = -6\)	\(f''(0) = 12\)
Sign of \(f''\)	\(-\)	\(+\)
\(f\)	concave down \(\cap\)	concave up \(\cup\)

Concave down on \((-\infty,-2)\), concave up on \((-2,\infty)\), and an inflection point at \(x = -2\).

Example Let \(f(x) = x^3 + 6x^2 - 15x + 2\) (same as above). Find the intervals of increasing and decreasing, all local extrema, the intervals of concavity, all inflection points, and use this to sketch a graph. \[ f'(x) = 3x^2+12x-15 = 3(x^2+4x-5) = 3(x-1)(x+5) = 0 \;\Longrightarrow\; x = 1,\ x = -5. \]

Interval	\((-\infty,-5)\)	\((-5,1)\)	\((1,\infty)\)
Test point	\(f'(-6) = 3(-7)(-1) > 0\)	\(f'(0) = -15 < 0\)	\(f'(2) = 3(1)(7) > 0\)
\(f\)	increasing ↗	decreasing ↘	increasing ↗

Increasing on \((-\infty,-5)\cup(1,\infty)\); decreasing on \((-5,1)\). Plug the critical points into the original function: local max at \(x=-5\) with \(f(-5) = -125+150+75+2 = 102\); local min at \(x = 1\) with \(f(1) = 1+6-15+2 = -6\).

Concave down on \((-\infty,-2)\), concave up on \((-2,\infty)\). Inflection point at \(x=-2\), where \(f(-2) = -8+24+30+2 = 48\).

Coming from the lower left: increasing and concave down up to the peak \((-5,102)\); decreasing and still concave down until the inflection point \((-2,48)\); decreasing but now concave up down to the valley \((1,-6)\); then increasing and concave up.

Example (Sketching from Information) Sketch a graph of a function \(f\) that is increasing on \((-\infty,2)\) and \((4,\infty)\), decreasing on \((2,4)\), concave down on \((-\infty,3)\) and concave up on \((3,\infty)\).

Combine the two charts: the graph rises while bending downward until a peak (local max) at \(x = 2\); falls, still bending down, until \(x=3\), where it switches to bending up (inflection point); keeps falling, now bending up, to a valley (local min) at \(x = 4\); then rises, bending up. Any curve with this shape is a correct sketch — the exact heights are not determined by this information.

Example The highest power in the denominator is \(x^2\): \[ \lim_{x\to\pm\infty}\frac{3x^2+4x}{5x^2-3x} = \lim_{x\to\pm\infty}\frac{\frac{3x^2}{x^2}+\frac{4x}{x^2}}{\frac{5x^2}{x^2}-\frac{3x}{x^2}} = \lim_{x\to\pm\infty}\frac{3+\frac4x}{5-\frac3x} = \frac{3+0}{5-0} = \frac35 = 0.6. \]

Example Divide each term by \(x\) (the highest power in the denominator): \[ \lim_{x\to\infty}\frac{x^2+4x}{3x+1} = \lim_{x\to\infty}\frac{\frac{x^2}{x}+\frac{4x}{x}}{\frac{3x}{x}+\frac1x} = \lim_{x\to\infty}\frac{x+4}{3+\frac1x} = \infty, \]

since the top grows without bound while the bottom approaches 3. Similarly \(\displaystyle\lim_{x\to-\infty}\frac{x+4}{3+\frac1x} = -\infty\).

Example Divide each term by \(x^4\): \[ \lim_{x\to\infty}\frac{3x^3+6x}{4x^4+5} = \lim_{x\to\infty}\frac{\frac3x+\frac{6}{x^3}}{4+\frac{5}{x^4}} = \frac{0+0}{4+0} = 0. \]

Example Use limits to determine if each function has a horizontal asymptote.

• \(f(x) = \dfrac{x^2+4x}{3x^2+2}\): dividing by \(x^2\), \(\displaystyle\lim_{x\to\pm\infty}\frac{1+\frac4x}{3+\frac{2}{x^2}} = \frac13\). So \(y = \tfrac13\) is a horizontal asymptote.
• \(g(x) = \dfrac{x^2+3x}{x-5}\): dividing by \(x\), \(\displaystyle\lim_{x\to\pm\infty}\frac{x+3}{1-\frac5x} = \pm\infty\). No horizontal asymptote.
• \(h(x) = \dfrac{x^3+6x}{x^4-6x}\): dividing by \(x^4\), \(\displaystyle\lim_{x\to\pm\infty}\frac{\frac1x+\frac{6}{x^3}}{1-\frac{6}{x^3}} = \frac01 = 0\). So \(y = 0\) is a horizontal asymptote.

Example Use all the steps to sketch a graph of \(f(x) = \dfrac{x^2}{1-x^2}\).

\(1 - x^2 = (1+x)(1-x)\ne 0\), so \(x\ne\pm1\). Domain: \((-\infty,-1)\cup(-1,1)\cup(1,\infty)\).

\(y\)-intercept: \(f(0) = \frac01 = 0\). \(x\)-intercepts: \(\frac{x^2}{1-x^2} = 0\) means \(x^2 = 0\), so \(x = 0\). The only intercept is the origin.

\[ \lim_{x\to\pm\infty}\frac{x^2}{1-x^2} = \lim_{x\to\pm\infty}\frac{\frac{x^2}{x^2}}{\frac{1}{x^2}-\frac{x^2}{x^2}} = \lim_{x\to\pm\infty}\frac{1}{\frac1{x^2}-1} = \frac{1}{0-1} = -1. \]

Horizontal asymptote at \(y = -1\).

At \(x = \pm1\) the top is \(1\) and the bottom is \(0\) ("\(\frac10\)"), so the function blows up. Checking signs of the bottom just to each side:

\[ \lim_{x\to1^-}\frac{x^2}{1-x^2} = +\infty,\quad \lim_{x\to1^+}\frac{x^2}{1-x^2} = -\infty,\quad \lim_{x\to-1^-}\frac{x^2}{1-x^2} = -\infty,\quad \lim_{x\to-1^+}\frac{x^2}{1-x^2} = +\infty. \]

Vertical asymptotes at \(x = -1\) and \(x = 1\).

\[ f'(x) = \frac{(1-x^2)(2x) - x^2(-2x)}{(1-x^2)^2} = \frac{2x - 2x^3 + 2x^3}{(1-x^2)^2} = \frac{2x}{(1-x^2)^2}. \]

Critical point: \(x = 0\). (\(x = \pm1\) are not in the domain but still go on the number line.) The denominator is always positive, so the sign of \(f'\) is the sign of \(2x\):

Interval	\((-\infty,-1)\)	\((-1,0)\)	\((0,1)\)	\((1,\infty)\)
Sign of \(f'\)	\(-\)	\(-\)	\(+\)	\(+\)
\(f\)	decreasing	decreasing	increasing	increasing

Decreasing on \((-\infty,-1)\cup(-1,0)\), increasing on \((0,1)\cup(1,\infty)\), local min at \(x = 0\).

Quotient rule again (the chain rule gives \(\frac{d}{dx}(1-x^2)^2 = 2(1-x^2)(-2x)\)):

\[ f''(x) = \frac{(1-x^2)^2(2) - 2x\cdot 2(1-x^2)(-2x)}{(1-x^2)^4} = \frac{(1-x^2)\bigl[2(1-x^2) + 8x^2\bigr]}{(1-x^2)^4} = \frac{2+6x^2}{(1-x^2)^3}. \]

The top \(2+6x^2\) is never zero (it is always positive). The bottom is zero at \(x = \pm1\). So the sign of \(f''\) is the sign of \((1-x^2)^3\), which is the sign of \(1-x^2\):

Interval	\((-\infty,-1)\)	\((-1,1)\)	\((1,\infty)\)
Sign of \(f''\)	\(-\)	\(+\)	\(-\)
\(f\)	concave down	concave up	concave down

The concavity changes at \(x = \pm1\), but these are not inflection points because they are not in the domain.

Example (Sketching from Properties) A function satisfies the following. Use this to sketch a graph.

• Domain \((-\infty,2)\cup(2,\infty)\), and \(\lim_{x\to2^+}f(x) = \lim_{x\to2^-}f(x) = \infty\).
• Horizontal asymptotes at \(y = 4\) and \(y = -2\).
• \(y\)-intercept of 5 and \(x\)-intercept of 4.
• \(f'(x)>0\) on \((-\infty,2)\) and \(f'(x) < 0\) on \((2,\infty)\).
• \(f''(x) > 0\) on \((-\infty,2)\) and \((2,\infty)\).

Draw the vertical asymptote \(x = 2\) and the horizontal lines \(y = 4\) and \(y = -2\). On the left piece, \(f\) is increasing and concave up and shoots up to \(\infty\) at \(x=2\); since it is increasing, it must level off toward \(y = 4\) as \(x\to-\infty\) (staying above it), passing through \((0,5)\). On the right piece, \(f\) comes down from \(\infty\) near \(x = 2\), decreasing and concave up, crosses the \(x\)-axis at \(x=4\), and levels off toward \(y=-2\) as \(x\to\infty\).

Example (Fencing a Garden) We have 100 feet of fencing and want to fence in a rectangular garden along a rock wall (no fence is needed along the wall). What dimensions maximize the area of the garden?

Objective: maximize the area \(A = \ell\cdot w\).

Constraint: the fence covers two sides of length \(w\) and one side of length \(\ell\): \(2w + \ell = 100\), so \(\ell = 100-2w\).

One variable + domain:

\[ A = (100-2w)\,w = 100w - 2w^2, \qquad 0\le w\le 50. \]

(The width can't be negative, and \(\ell = 100 - 2w\) can't be negative.)

Derivative:

\[ A' = 100 - 4w = 0 \;\Longrightarrow\; w = 25. \]

Check:

\(w\)	\(A = (100-2w)w\)
0	0
25	\((100-50)(25) = 1250\)	absolute max
50	0

The dimensions are \(w = 25\) ft and \(\ell = 100-2(25) = 50\) ft, for a maximum area of 1250 ft².

Example (Maximizing the Volume of a Box) An open-top box is made from a 24 in \(\times\) 36 in piece of cardboard by removing a square from each corner and then folding up the sides. What is the maximum volume of the box?

Variables: let \(x\) be the side length of each removed square. Folding up the flaps makes a box of height \(x\); each side of the base loses \(x\) at both ends.

Objective: maximize volume \(V = \ell\cdot w\cdot h\).

Constraint: \(\ell = 24-2x\), \(w = 36-2x\), \(h = x\).

One variable + domain:

\[ V = x(24-2x)(36-2x) = 4x^3 - 120x^2 + 864x, \qquad 0\le x\le 12. \]

(We can't cut out more than half of the 24-inch side.)

Derivative:

\[ V' = 12x^2 - 240x + 864 = 12(x^2 - 20x + 72) = 0. \]

Try to factor; if that fails, use the quadratic formula:

\[ x = \frac{20\pm\sqrt{400 - 4(1)(72)}}{2} = \frac{20\pm\sqrt{112}}{2} = 10\pm2\sqrt7. \]

\(10+2\sqrt7\approx 15.29\) is outside \([0,12]\), so the only critical point is \(x = 10-2\sqrt7\approx 4.708\).

Check:

\(x\)	\(V = x(24-2x)(36-2x)\)
0	0
\(10-2\sqrt7\approx4.708\)	\(\approx 1825.3\)	max
12	0

(A sign chart agrees: \(V'\) is \(+\) on \((0, 4.708)\) and \(-\) on \((4.708, 12)\).) The maximum volume is about 1825 in³, when squares of side \(10-2\sqrt7\approx 4.71\) in are removed.

Example (Maximizing Revenue) A car rental company is willing to rent cars for between $50 and $200 a day. If they price cars at $50, they can rent 750, and for each $1 they increase the price, the demand goes down by 5 cars. What price maximizes revenue?

Let \(x\) be the number of $1 increases. Making a table helps see the pattern:

Price	$50	$51	$52	$53	…	$60	…	\(50+x\)
# cars	750	745	740	735	…	\(750-5(10)\)	…	\(750-5x\)

Revenue \(=(\#\text{ items})(\text{price per item})\):

\[ R(x) = (750-5x)(50+x) = 37500 + 500x - 5x^2, \qquad 0\le x\le 150. \]

(Price \(50+x\) must be at most $200.)

\[ R'(x) = 500 - 10x = 0 \;\Longrightarrow\; x = 50. \]

\(x\)	\(R(x) = (750-5x)(50+x)\)
0	\(750\cdot 50 = 37{,}500\)
50	\(500\cdot100 = 50{,}000\)	max
150	\(0\cdot200 = 0\)

The revenue-maximizing price is \(50+x = 50+50 = \)$100 per day (renting 500 cars for $50,000).

Example (Minimizing Cost) A company wants to produce a sturdy open-top rectangular container with a square base whose volume is 128 ft³. The sides cost $2 per square foot and the bottom costs $6 per square foot. Find the dimensions that minimize cost.

Variables: let \(x\) be the side length of the square base and \(y\) the height.

Objective: minimize cost.

\[ C = (\text{cost per ft}^2)(\text{area of bottom}) + (\text{cost per ft}^2)(\text{area of a side})(\#\text{ sides}) = 6x^2 + 2(xy)(4) = 6x^2+8xy. \]

Constraint: \(V = x^2y = 128\), so \(y = \dfrac{128}{x^2}\).

One variable + domain:

\[ C = 6x^2 + 8x\cdot\frac{128}{x^2} = 6x^2 + \frac{1024}{x}, \qquad x > 0. \]

Here \(x > 0\) (if \(x = 0\) the volume would be 0), but there is no upper bound, so we can't use the closed interval table. Instead we use a sign chart.

Derivative:

\[ C'(x) = 12x - \frac{1024}{x^2} = 0 \;\Longrightarrow\; 12x = \frac{1024}{x^2} \;\Longrightarrow\; 12x^3 = 1024 \;\Longrightarrow\; x^3 = \frac{256}{3} \;\Longrightarrow\; x = \sqrt[3]{\frac{256}{3}}\approx 4.40. \]

(\(C'\) is undefined at \(x = 0\), but that is not in the domain.)

Check: \(C'(1) = 12 - 1024 < 0\) and \(C'(10) = 120 - 10.24 > 0\). So \(C\) is decreasing before \(x\approx 4.40\) and increasing after: an absolute minimum at \(x = \sqrt[3]{256/3}\).

\[ y = \frac{128}{x^2} = \frac{128}{(256/3)^{2/3}}\approx 6.60. \]

The cheapest container has a base about 4.40 ft by 4.40 ft and a height of about 6.60 ft (costing about $348.89).