Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Optimization Problems

An optimization problem asks for the biggest or smallest possible value of something: the most area, the least material, the shortest distance. Derivatives find it, because a maximum or minimum sits where the slope is zero or at an end of the interval.

How It Works

The lesson's method, in order. Every step earns marks on a test, so write them all down.

1. Name what you are optimizing. "Maximize the area $A$ ." Draw a diagram and label it.

2. Write a formula for it. Usually it has two variables, like $A = xy$ .

3. Use the constraint to get one variable. The problem always gives a fixed amount: a perimeter, a volume, a curve the point must sit on. Solve that for one variable and substitute, so the formula has only one.

4. State the domain. Lengths are positive, and you cannot cut more than the sheet. The domain gives you the endpoints to check.

5. Differentiate, set the derivative to zero, solve, and verify. Compare the function at each critical number and at the endpoints, or use the second derivative test: $f''> 0$ is a minimum and $f'' < 0$ is a maximum.

Then answer the question that was asked. If it asks for the dimensions, give both, with units. A bare $x = 60$ is not an answer.

Where students lose marks: skipping the domain, so a critical number outside it sneaks into the answer; never checking it is a max rather than a min; and keeping two variables, so there is nothing to differentiate.

A shortcut for distance: to find the point on a curve closest to another point, minimize the distance squared. It has the same minimum, and it gets rid of the square root before you differentiate.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Two positive numbers add to $24$ . Find them so that their product is as large as possible.
Show solution
1. Maximize $P = xy$ with $x + y = 24$ , so $y = 24 - x$ .
2. $P = x(24 - x) = 24x - x^2$ on $0 < x < 24$ .
3. $P' = 24 - 2x = 0$ gives $x = 12$ .
4. $P'' = -2 < 0$ , so it is a maximum. $y = 12$ and $P = 144$ .
Answer: $12$ and $12$ , with product $144$ .
Easy
A farmer has $240$ m of fence for a rectangular pen along a river. The river side needs no fence. Find the dimensions with the largest area.
Show solution
1. Let $x$ be each side perpendicular to the river and $y$ the side parallel to it. Fence: $2x + y = 240$ , so $y = 240 - 2x$ .
2. $A = xy = x(240 - 2x) = 240x - 2x^2$ on $0 \le x \le 120$ .
3. $A' = 240 - 4x = 0$ gives $x = 60$ .
4. Check: $A(0) = 0$ , $A(120) = 0$ , $A(60) = 60 \times 120 = 7200$ . The maximum is at $x = 60$ , where $y = 120$ .
Answer: $60$ m by $120$ m (the $120$ m side along the river), for an area of $7200$ m².
Medium
An open box is made from a $24$ cm by $24$ cm sheet by cutting equal squares from the corners and folding up the sides. Find the largest possible volume.
Show solution
1. Let $x$ be the side of each cut square. The base is $24 - 2x$ on each side and the height is $x$ .
2. $V = x(24 - 2x)^2$ on $0 \le x \le 12$ .
3. Product and chain rule: $V' = (24 - 2x)^2 - 4x(24 - 2x) = (24 - 2x)(24 - 6x)$ .
4. $V' = 0$ at $x = 12$ or $x = 4$ . At $x = 12$ the base vanishes.
5. Check: $V(0) = 0$ , $V(12) = 0$ , $V(4) = 4 \times 16^2 = 1024$ .
Answer: Cut $4$ cm squares, for a maximum volume of $1024$ cm³.
Medium
A box with a square base and no top must hold $32\,000$ cm³. Find the dimensions that use the least material.
Show solution
1. Let the base be $x$ by $x$ and the height $h$ . Minimize the surface area $S = x^2 + 4xh$ .
2. Volume constraint: $x^2h = 32\,000$ , so $h = \frac{32\,000}{x^2}$ .
3. Substitute: $S = x^2 + \frac{128\,000}{x}$ for $x > 0$ .
4. $S' = 2x - \frac{128\,000}{x^2} = 0$ gives $x^3 = 64\,000$ , so $x = 40$ .
5. $S'' = 2 + \frac{256\,000}{x^3} > 0$ , so it is a minimum. $h = \frac{32\,000}{1600} = 20$ , and $S = 1600 + 3200 = 4800$ .
Answer: Base $40$ cm by $40$ cm, height $20$ cm, using $4800$ cm² of material.
Test level
Find the points on the parabola $y = x^2$ closest to the point $(0, 2)$ .
Show solution
1. A point on the curve is $(x, x^2)$ . Minimize the distance squared: $D = x^2 + (x^2 - 2)^2$ .
2. $D' = 2x + 2(x^2 - 2)(2x) = 2x(2x^2 - 3)$ .
3. $D' = 0$ at $x = 0$ or $x^2 = \frac{3}{2}$ , so $x = \pm\frac{\sqrt{6}}{2}$ .
4. Compare: $D(0) = 4$ , while $D\left(\pm\frac{\sqrt{6}}{2}\right) = \frac{3}{2} + \frac{1}{4} = \frac{7}{4}$ . So $x = 0$ is a local maximum of $D$ , a trap, and the minimum is at $x = \pm\frac{\sqrt{6}}{2}$ .
5. There $y = x^2 = \frac{3}{2}$ , and the distance is $\sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2}$ .
Answer: $\left(\frac{\sqrt{6}}{2}, \frac{3}{2}\right)$ and $\left(-\frac{\sqrt{6}}{2}, \frac{3}{2}\right)$ , at distance $\frac{\sqrt{7}}{2}$ .

Learn It on Martian Lab

Martian Lab teaches this in the Optimization 1 lesson of its MCV4U Optimization unit. Watch it, then take the Optimization practice tests to see if it stuck.

Watch the Optimization 1 lesson free

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Free Optimization practice test, with solutions

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