Optimisation with Calculus

Mathematics · Calculus & Analysis · ages 17-18

Name ______________________ Date ____________

If a question asks for the largest possible area, the value of x on its own is not a complete answer.
Circle one: True False
After solving the derivative equal to zero, you should check what kind of stationary point you have found.
Circle one: True False
For A = 12x - x², at what value of x is dA/dx zero?
Answer: ______________
A rectangle with sides x and y has a perimeter of 20, so 2x + 2y = 20. What is y when x = 3?
Answer: ______________
A rectangle is made from 24 m of fencing on all four sides. What dimensions give the largest area?
- 4 m by 8 m
- 6 m by 6 m
- 3 m by 9 m
- 12 m by 12 m
Two positive numbers add up to 20. What pair makes their product as large as possible?
- 1 and 19
- 10 and 10
- 5 and 15
- 2 and 18
Kim writes the area as A = xy, differentiates straight away and gets dA/dx = y. Where is the slip?
- the area of a rectangle is not xy
- she should have differentiated with respect to y instead
- the answer should have been dA/dx = x
- y is not a fixed number here, so the constraint has to replace it first
A farmer has 60 m of fencing for three sides of a rectangle, with a wall as the fourth. The two equal sides are x. At what value of x is the area largest?
Answer: ______________
A rectangle has an area of 64. Its perimeter is P = 2x + 128/x. What is the smallest perimeter?
Answer: ______________
Raj maximises A = 30x - x², finds x = 15 correctly, and reports the largest area as 15. Where is the slip?
- 15 is the value of x, and the area is A(15), which is 225
- the stationary point is at x = 30, not x = 15
- he should have used the second derivative to find x
- the largest area is 30, from the number in front of the x

Answer key

For grown-ups. Fold this page away before handing over the rest.

Optimisation with Calculus W1-mt_9I0ql39Lzw-s1

True · x is the side length that produces the largest area, not the area itself. Substituting it back into the area function is the last step.
True · A stationary point is not automatically the kind you wanted. The second derivative settles it, and the check is part of the answer.
6 · dA/dx = 12 - 2x. Setting that to zero gives 2x = 12, so x = 6.
7 · Rearranging gives 2y = 20 - 2x, so y = 10 - x. With x = 3 that is 7.
6 m by 6 m · From 2x + 2y = 24 you get y = 12 - x, so A = x(12 - x) = 12x - x². Then dA/dx = 12 - 2x is zero at x = 6, and y = 6 as well.
10 and 10 · With x + y = 20 the product is P = x(20 - x) = 20x - x². Then dP/dx = 20 - 2x is zero at x = 10, so both numbers are 10 and the product is 100.
y is not a fixed number here, so the constraint has to replace it first · Treating y as a constant would only be right if it were one. Here y depends on x through the constraint, so it has to be substituted out before differentiating.
15 · The constraint is 2x + y = 60, so y = 60 - 2x and A = x(60 - 2x) = 60x - 2x². Then dA/dx = 60 - 4x is zero at x = 15.
32 · dP/dx = 2 - 128/x² is zero when x² = 64, so x = 8. Then P = 16 + 16 = 32.
15 is the value of x, and the area is A(15), which is 225 · His calculus is right. The value of x that makes the area largest is 15, but the area itself is 30 times 15 minus 15², which is 450 - 225 = 225.