Practice Questions: Applications of Differential Calculus (Optimisation)

$$\begin{align*} \text{Let the numbers be } a \text{ and } b. \\ \text{Constraint: } a + b &= 10 \\ b &= 10 - a \\ \\ \text{Product: } P &= a \cdot b^2 \\ P(a) &= a(10 - a)^2 \\ &= a(100 - 20a + a^2) \\ &= 100a - 20a^2 + a^3 \\ \\ \text{Differentiate:} \\ P'(a) &= 100 - 40a + 3a^2 \\ &= 3a^2 - 40a + 100 \\ \\ \text{Set } P'(a) = 0: \\ 3a^2 - 40a + 100 &= 0 \\ \text{Using quadratic formula:} \\ a &= \frac{40 \pm \sqrt{1600 - 1200}}{6} \\ &= \frac{40 \pm 20}{6} \\ a &= 10 \text{ or } a = \frac{10}{3} \\ \\ \text{If } a = 10, \text{ then } b = 0 \text{ (not allowed)} \\ \text{If } a = \frac{10}{3}, \text{ then } b = \frac{20}{3} \\ \\ \text{Check using second derivative:} \\ P''(a) &= 6a - 40 \\ P''\left(\frac{10}{3}\right) &= 20 - 40 = -20 < 0 \quad \text{(maximum)} \end{align*}$$

Answer: The numbers are $a = \frac{10}{3}$ and $b = \frac{20}{3}$ (or $3\frac{1}{3}$ and $6\frac{2}{3}$).

$$\begin{align*} \text{Constraint (perimeter):} \\ 2x + 2y &= 30 \\ y &= 15 - x \\ \\ \text{Area (L-shaped):} \\ A &= 2xy - x^2 \\ A(x) &= 2x(15 - x) - x^2 \\ &= 30x - 2x^2 - x^2 \\ &= 30x - 3x^2 \\ \\ \text{Differentiate:} \\ A'(x) &= 30 - 6x \\ \\ \text{Set } A'(x) = 0: \\ 30 - 6x &= 0 \\ x &= 5 \text{ m} \\ \\ \text{Then: } y &= 15 - 5 = 10 \text{ m} \\ \\ \text{Check:} \\ A''(x) &= -6 < 0 \quad \text{(maximum)} \end{align*}$$

Answer: $x = 5$ m and $y = 10$ m maximise the area.

$$\begin{align*} \text{a) Volume constraint:} \\ x^2h &= 750 \\ h &= \frac{750}{x^2} \\ \\ \text{Surface area:} \\ A &= \text{base} + \text{top} + \text{four sides} \\ &= x^2 + 2x^2 + 4xh \\ &= 3x^2 + 4xh \\ &= 3x^2 + 4x\left(\frac{750}{x^2}\right) \\ &= 3x^2 + \frac{3000}{x} \quad \checkmark \\ \\ \text{b) Minimize:} \\ A'(x) &= 6x - \frac{3000}{x^2} \\ \\ \text{Set } A'(x) = 0: \\ 6x - \frac{3000}{x^2} &= 0 \\ 6x &= \frac{3000}{x^2} \\ 6x^3 &= 3000 \\ x^3 &= 500 \\ x &= \sqrt[3]{500} \approx 7.94 \text{ cm} \\ \\ \text{Check:} \\ A''(x) &= 6 + \frac{6000}{x^3} \\ A''(\sqrt[3]{500}) &= 6 + \frac{6000}{500} > 0 \quad \text{(minimum)} \end{align*}$$

Answers:

• a) $A(x) = \frac{3000}{x} + 3x^2$ ✓
• b) $x = \sqrt[3]{500} \approx 7.94$ cm minimises surface area

$$\begin{align*} \text{Let } l &= \text{length parallel to wall} \\ \text{Let } w &= \text{width perpendicular to wall} \\ \\ \text{Constraint:} \\ w + 2l &= 160 \\ w &= 160 - 2l \\ \\ \text{Area:} \\ A(l) &= l \cdot w \\ &= l(160 - 2l) \\ &= 160l - 2l^2 \\ \\ \text{Differentiate:} \\ A'(l) &= 160 - 4l \\ \\ \text{Set } A'(l) = 0: \\ 160 - 4l &= 0 \\ l &= 40 \text{ m} \\ \\ \text{Then: } w &= 160 - 2(40) = 80 \text{ m} \\ \\ \text{Check:} \\ A''(l) &= -4 < 0 \quad \text{(maximum)} \\ \\ \text{Maximum area:} \\ A &= 40 \times 80 = 3200 \text{ m}^2 \end{align*}$$

Answer: Dimensions are $40$ m (parallel to wall) by $80$ m (perpendicular to wall), giving maximum area of $3200$ m².

$$\begin{align*} \text{Let } l &= \text{length}, w = \text{width} \\ \\ \text{Fencing needed:} \\ 2l + 3w &= 400 \text{ (two lengths + three widths)} \\ l &= \frac{400 - 3w}{2} \\ \\ \text{Area:} \\ A(w) &= l \cdot w \\ &= \frac{400 - 3w}{2} \cdot w \\ &= \frac{400w - 3w^2}{2} \\ &= 200w - \frac{3w^2}{2} \\ \\ \text{Differentiate:} \\ A'(w) &= 200 - 3w \\ \\ \text{Set } A'(w) = 0: \\ 200 - 3w &= 0 \\ w &= \frac{200}{3} \approx 66.67 \text{ m} \\ \\ \text{Then: } l &= \frac{400 - 3(200/3)}{2} = \frac{200}{2} = 100 \text{ m} \\ \\ \text{Check:} \\ A''(w) &= -3 < 0 \quad \text{(maximum)} \end{align*}$$

Answer: Dimensions are $100$ m by $\frac{200}{3}$ m (or $66.67$ m).

$$\begin{align*} \text{Volume constraint:} \\ \pi r^2h &= 1000 \\ h &= \frac{1000}{\pi r^2} \\ \\ \text{Surface area:} \\ A &= 2\pi r^2 + 2\pi rh \\ &= 2\pi r^2 + 2\pi r\left(\frac{1000}{\pi r^2}\right) \\ &= 2\pi r^2 + \frac{2000}{r} \\ \\ \text{Differentiate:} \\ A'(r) &= 4\pi r - \frac{2000}{r^2} \\ \\ \text{Set } A'(r) = 0: \\ 4\pi r - \frac{2000}{r^2} &= 0 \\ 4\pi r &= \frac{2000}{r^2} \\ 4\pi r^3 &= 2000 \\ r^3 &= \frac{500}{\pi} \\ r &= \sqrt[3]{\frac{500}{\pi}} \approx 5.42 \text{ cm} \\ \\ \text{Height:} \\ h &= \frac{1000}{\pi(5.42)^2} \approx 10.84 \text{ cm} \\ \\ \text{Check:} \\ A''(r) &= 4\pi + \frac{4000}{r^3} > 0 \quad \text{(minimum)} \end{align*}$$

Answer: $r \approx 5.42$ cm and $h \approx 10.84$ cm (note: $h = 2r$).

$$\begin{align*} \text{Let } x &= \text{width of printed area} \\ \text{Let } y &= \text{height of printed area} \\ \\ \text{Constraint:} \\ xy &= 180 \\ y &= \frac{180}{x} \\ \\ \text{Total dimensions:} \\ \text{Width} &= x + 6 \text{ (margins of 3 cm each side)} \\ \text{Height} &= y + 10 \text{ (margins of 5 cm top and bottom)} \\ \\ \text{Total area:} \\ A &= (x + 6)(y + 10) \\ &= (x + 6)\left(\frac{180}{x} + 10\right) \\ &= 180 + 10x + \frac{1080}{x} + 60 \\ &= 10x + \frac{1080}{x} + 240 \\ \\ \text{Differentiate:} \\ A'(x) &= 10 - \frac{1080}{x^2} \\ \\ \text{Set } A'(x) = 0: \\ 10 - \frac{1080}{x^2} &= 0 \\ 10x^2 &= 1080 \\ x^2 &= 108 \\ x &= \sqrt{108} = 6\sqrt{3} \approx 10.39 \text{ cm} \\ \\ \text{Then: } y &= \frac{180}{6\sqrt{3}} = \frac{30}{\sqrt{3}} = 10\sqrt{3} \approx 17.32 \text{ cm} \\ \\ \text{Poster dimensions:} \\ \text{Width} &= 6\sqrt{3} + 6 \approx 16.39 \text{ cm} \\ \text{Height} &= 10\sqrt{3} + 10 \approx 27.32 \text{ cm} \end{align*}$$

Answer: Poster should be approximately $16.39$ cm by $27.32$ cm.

$$\begin{align*} \text{Differentiate:} \\ f'(v) &= \frac{3}{40}v - 6 \\ \\ \text{Set } f'(v) = 0: \\ \frac{3}{40}v - 6 &= 0 \\ \frac{3}{40}v &= 6 \\ v &= 6 \times \frac{40}{3} \\ v &= 80 \text{ km/h} \\ \\ \text{Check:} \\ f''(v) &= \frac{3}{40} > 0 \quad \text{(minimum)} \end{align*}$$

Answer: The most economical cruising speed is $80$ km/h.

$$\begin{align*} \text{Volume constraint:} \\ \pi r^2h &= 500\pi \\ h &= \frac{500}{r^2} \\ \\ \text{Surface area (no top):} \\ A &= \pi r^2 + 2\pi rh \\ &= \pi r^2 + 2\pi r\left(\frac{500}{r^2}\right) \\ &= \pi r^2 + \frac{1000\pi}{r} \\ \\ \text{Differentiate:} \\ A'(r) &= 2\pi r - \frac{1000\pi}{r^2} \\ \\ \text{Set } A'(r) = 0: \\ 2\pi r - \frac{1000\pi}{r^2} &= 0 \\ 2\pi r &= \frac{1000\pi}{r^2} \\ 2r^3 &= 1000 \\ r^3 &= 500 \\ r &= \sqrt[3]{500} \approx 7.94 \text{ cm} \\ \\ \text{Check:} \\ A''(r) &= 2\pi + \frac{2000\pi}{r^3} > 0 \quad \text{(minimum)} \end{align*}$$

Answer: $r = \sqrt[3]{500} \approx 7.94$ cm minimises the material used.

$$\begin{align*} \text{Let } x &= \text{length for circle} \\ \text{Then } 100 - x &= \text{length for square} \\ \\ \text{Circle:} \\ 2\pi r &= x \\ r &= \frac{x}{2\pi} \\ A_{\text{circle}} &= \pi r^2 = \pi\left(\frac{x}{2\pi}\right)^2 = \frac{x^2}{4\pi} \\ \\ \text{Square:} \\ 4s &= 100 - x \\ s &= \frac{100 - x}{4} \\ A_{\text{square}} &= s^2 = \left(\frac{100 - x}{4}\right)^2 = \frac{(100 - x)^2}{16} \\ \\ \text{Total area:} \\ A(x) &= \frac{x^2}{4\pi} + \frac{(100 - x)^2}{16} \\ \\ \text{Differentiate:} \\ A'(x) &= \frac{2x}{4\pi} + \frac{2(100 - x)(-1)}{16} \\ &= \frac{x}{2\pi} - \frac{100 - x}{8} \\ \\ \text{Set } A'(x) = 0: \\ \frac{x}{2\pi} &= \frac{100 - x}{8} \\ 8x &= 2\pi(100 - x) \\ 8x &= 200\pi - 2\pi x \\ 8x + 2\pi x &= 200\pi \\ x(8 + 2\pi) &= 200\pi \\ x &= \frac{200\pi}{8 + 2\pi} \approx 56 \text{ cm} \end{align*}$$

Answer: Cut the wire at approximately $56$ cm from one end (for the circle), leaving $44$ cm for the square.

$$\begin{align*} \text{Let } x &= \text{width of each corral} \\ \text{Let } y &= \text{length of each corral} \\ \\ \text{Fencing needed:} \\ 3x + 2y &= 600 \text{ (three widths, two lengths)} \\ y &= \frac{600 - 3x}{2} \\ \\ \text{Total area:} \\ A &= 2xy \text{ (two corrals)} \\ A(x) &= 2x\left(\frac{600 - 3x}{2}\right) \\ &= x(600 - 3x) \\ &= 600x - 3x^2 \\ \\ \text{Differentiate:} \\ A'(x) &= 600 - 6x \\ \\ \text{Set } A'(x) = 0: \\ 600 - 6x &= 0 \\ x &= 100 \text{ m} \\ \\ \text{Then: } y &= \frac{600 - 300}{2} = 150 \text{ m} \\ \\ \text{Check:} \\ A''(x) &= -6 < 0 \quad \text{(maximum)} \\ \\ \text{Maximum total area:} \\ A &= 2(100)(150) = 30\,000 \text{ m}^2 \end{align*}$$

Answer: Each corral should be $100$ m by $150$ m, for a total area of $30\,000$ m².

$$\begin{align*} \text{Differentiate:} \\ P'(x) &= -4x + 120 \\ \\ \text{Set } P'(x) = 0: \\ -4x + 120 &= 0 \\ x &= 30 \\ \\ \text{Check:} \\ P''(x) &= -4 < 0 \quad \text{(maximum)} \\ \\ \text{Maximum profit:} \\ P(30) &= -2(30)^2 + 120(30) - 1000 \\ &= -1800 + 3600 - 1000 \\ &= 800 \end{align*}$$

Answer: Produce $30 \times 100 = 3000$ units to maximise profit at R$800$ (in hundreds).