A box has a square base of side x and height y. (assume the box does not have a top)?

Question

a) Find the dimension x, y for which the volume is 12 and the surface area is as small as possible.

b) Find the dimensions for which the surface area is 20 and the volume is as large as possible.

Answer 1

Volume: V = y*x^2

Surface Area: A = x^2 + 4*x*y

12 = y*x^2

y = 12 / (x^2)

A = x^2 + 4*x*(12 / x^2) = x^2 + 48 / x

dA / dx = 0 = 2x - 48 / x^2

2x = 48 / x^2

2x^3 = 48

x^3 = 24

x = 2 sqrt(3)

Answer 2

a) Surface area is (height = h):

(base) x²

(4 sides) 4hx

A = x² + 4hx

V = 12 = hx²

h = 12/x²

substituting

A = x² + 4x(12/x²)

A = x² + 48/x

dA/dx = 2x - 48/x² = 0

assuming x not 0, mult by x²/2

x³ - 24 = 0

x = ³√24 = 2.88

h = 12/ 2.88² = 0.5

b)There is only one solution to the equation, there is no maximum.

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