MCV4U Applications of Derivatives Practice Test (with Solutions)

Free Optimization Practice Test

4 questions, exactly as a free account gets them in the first Optimization test on Martian Lab. Try each one on paper first, then open the answer.

1. You are given a piece of sheet metal that is twice as long as it is wide and has an area of 800 m$^2$. Find the dimensions of the rectangular box that would contain a maximum volume if it were constructed from this piece of metal by cutting out squares of equal area at all four corners and folding up the sides. The box does not have a lid. Include a diagram with your setup/solution.

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   Let $x$ be the width; $2x$ is the length. $2x \cdot x = 800$ $x^2 = 400$ $x = 20$ Let $y$ be the side length of the square that gets cut off. Maximize volume $V$. $V = (2x-2y)(x-2y)y$ $= (40-2y)(20-2y)y$ $= (800 - 120y + 4y^2)y$ $= 4y^3 - 120y^2 + 800y$, where $0 \le y \le 10$ $v' = 12y^2 - 240y + 800 = 0$ $3y^2 - 60y + 200 = 0$ $y \approx 4.2$ or $y = 15.7$; reject $y = 15.7$ since $0 \le y \le 10$. $V(0) = V(10) = 0$ $V(4.2) = 1540$ (maximum) $20 - 8.4 = 11.6$ $40 - 8.4 = 31.6$ $\therefore$ A box with dimensions $11.6\text{ m} \times 31.6\text{ m} \times 4.2\text{ m}$ would yield the maximum volume.
2. A ship is 25 kilometres north of a raft when it started sailing south at a rate of 4 km/h and the raft started to drift east at a speed of 5 km/h. Determine the shortest distance between the ship and the raft. Include a diagram in your setup/solution.

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   Let $t$ be the time in hours after the ship and raft start moving. Minimize distance $d$. $d = \sqrt{x^2+y^2}$ $x = 5t$ $y = 25 - 4t$ $0 \le t \le \dfrac{25}{4}$ $d = \sqrt{(5t)^2 + (25-4t)^2}$ $= \sqrt{625 - 200t + 41t^2}$ $d'(t) = \dfrac{1}{2}\left(41t^2 - 200t + 625\right)^{-\frac{1}{2}} \cdot (82t - 200) = 0$ $t = \dfrac{200}{82} = \dfrac{100}{41}$ $d(0) = 25$ $d(6.25) = 31.25$ $d\!\left(\dfrac{100}{41}\right) \approx 19.5$ $\therefore$ The minimum distance between the ship and the raft is about 19.5 km.
3. A tall building is 27 m away from an 8 m wall that is parallel to it. A long ladder on the ground is to go over the wall and lean onto the building as shown. Determine the length of the shortest ladder that is required.

   

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   Minimize the length of the ladder $L$. Divide the length $L$ into two parts as shown in the diagram. $L = x + y$ Use similar triangles: $\dfrac{y}{8} = \dfrac{x}{\sqrt{x^2 - 27^2}}$ $y = \dfrac{8x}{\sqrt{x^2-729}}$ $L = x + \dfrac{8x}{\sqrt{x^2-729}}$ $L'(x) = \dfrac{(x^2-729)^{3/2} - 5832}{(x^2-729)^{3/2}} = 0$ $(x^2 - 729)^{3/2} = 5832$ $x^2 - 729 = 324$ $x^2 = 1053$ $x \approx 32.4$ $L \approx 46.9$ $\therefore$ The minimum length of the ladder is about 46.9 m.
4. Find the dimensions of the largest squared-based prism that can fit inside a square-based pyramid that has a height of 12 m and a base width of 18 m.

   

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   Let $x$ be the base length of the square-based prism and let $h$ be its height. Maximize volume $V$ of the square-based prism. $V = x^2 \cdot h$ Use similar triangles: $\dfrac{h}{9 - x/2} = \dfrac{12}{9}$ $9h = 108 - 6x$ $h = 12 - \dfrac{2}{3}x$ $V = x^2\!\left(12 - \dfrac{2}{3}x\right) = 12x^2 - \dfrac{2}{3}x^3$, where $0 \le x \le 18$ $V'(x) = 24x - 2x^2$ $= 2x(12 - x)$ $V'(x) = 0$ when $x = 0$ or $x = 12$ $V(0) = V(18) = 0$ $V(12) = 576$ The maximum volume occurs when $x = 12$. $h = 12 - \dfrac{2}{3} \times 12 = 4$ $\therefore$ A prism with dimensions $12\text{ m} \times 12\text{ m} \times 4\text{ m}$ would have maximum volume.

This Unit on Martian Lab

The lessons cover optimization in two parts, then related rates.

1. Optimization 1
2. Optimization 2
3. Related Rates

Then 4 full Optimization practice tests to check yourself.

Optimization FAQ

Optimization Topics

One idea at a time: a short explanation, then five worked examples.

• Optimization Problems
• Related Rates

Start Practicing

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