A box with a square base and open top must have a volume of 470596 cm^3 . We wish to find the dimensions of the box that minimize the amount of material used.

Question

First, find a formula for the surface area of the box in terms of only xx, the length of one side of the square base.

[Hint: use the volume formula to express the height of the box in terms of xx.]

Simplify your formula as much as possible.

A(x)=_________________

Next, find the derivative, 

A'(x)=_________________

Now, calculate when the derivative equals zero, that is, when A'(x)=0. [Hint: multiply both sides by x2x2.]

A'(x)=0 when x=____________________

We next have to make sure that this value of x gives a minimum value for the surface area. Let's use the second derivative test. Find A"(x).

A"(x)=__________________________

Evaluate A"(x) at the x-value you gave above.

_______________________

NOTE: Since your last answer is positive, this means that the graph of A(x) is concave up around that value, so the zero of A'(x) must indicate a local minimum for A(x). (Your boss is happy now.)

Answer 1

I don't understand the "xx" so I'm just going to use "x" for the length of one side of the square base -- especially when later on it's A(x) etc. With an open top there's one base and 4 sides, A = x² + 4xh. V = base*height = x²h = 470,596, then h = 470,596/x². Write A(x) = x² + 4x(470,596/x²) = x² + 1,882,384 x^-1.

A'(x) = 2x - 1,882,384 x^-2. Set to zero and multiply by x²: 2x³ = 1,882,384, x³ = 941,192, and x = 98.

A''(x) = 2 + 3,764,768 x^-3 and A''(98) = 2 + 3,764,768 / 941,192 = 6.