Optimisation with Calculus · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Optimisation with Calculus

Mathematics · Calculus & Analysis · ages 17-18
Name ______________________   Date ____________

  1. A question says: a rectangle has an area of 36, and you must make its perimeter as small as possible. What is the constraint?

    • the perimeter being as small as possible
    • the rectangle having four sides
    • the area being fixed at 36
    • the sides being called x and y

  2. For A = 12x - x², at what value of x is dA/dx zero?

    Answer: ______________

  3. For the same 40-perimeter rectangle, what is the maximum possible area?

    Answer: ______________

  4. An optimisation question does not arrive as a rule to differentiate. What has to happen first?

    • build the function yourself from the words
    • check the second derivative of the constraint
    • differentiate the constraint
    • sketch the curve

  5. An open box with a square base has a fixed volume of 256, and its surface area is smallest when the base side is 8. What is the height then?

    Answer: ______________

  6. For A = 20x - x² the stationary point is at x = 10. What kind is it?

    • A maximum, because the second derivative is -2
    • A minimum, because the second derivative is -2
    • A point of inflection, because the derivative is zero
    • Nothing, because endpoints were not checked

  7. An open box has a square base of side 4 cm and a volume of 48 cm³. What is its height?

    • 12 cm
    • 4 cm
    • 3 cm
    • 2 cm

  8. Kim writes the area as A = xy, differentiates straight away and gets dA/dx = y. Where is the slip?

    • she should have written A = x + y for the area
    • she should first have differentiated everything 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

  9. A rectangle has an area of 64. Its perimeter is P = 2x + 128/x. What is the smallest perimeter?

    Answer: ______________

  10. 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
LightMySky · lightmysky.comW1-mt_9I0ql39Lzw-s1

Answer key

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

Optimisation with Calculus W1-mt_9I0ql39Lzw-s1

  1. the area being fixed at 36 · The constraint is what is fixed. Here the area is pinned at 36, which links x and y. The perimeter is the thing being optimised, not the constraint.
  2. 6 · dA/dx = 12 - 2x. Setting that to zero gives 2x = 12, so x = 6.
  3. 100 · A = 20(10) - 10² = 200 - 100 = 100.
  4. build the function yourself from the words · The words describe a situation, not a function. Turning them into a rule you can differentiate is the part of the work that comes before any calculus.
  5. 4 · Volume x²h = 256 with x = 8 gives 64h = 256, so h = 4.
  6. A maximum, because the second derivative is -2 · A'' = -2 is negative, so the stationary point is a maximum.
  7. 3 cm · The constraint is x²h = 48, so h = 48 divided by 16, which is 3 cm.
  8. 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.
  9. 32 · dP/dx = 2 - 128/x² is zero when x² = 64, so x = 8. Then P = 16 + 16 = 32.
  10. 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.
Worksheet · LightMySky