# A box with a square base has no top. If 58 cm^2 of material is used what is the max possible volume for the box? need help urgently??

### 1 Answer | Add Yours

Let w, l, h be the dimensions of the box

The volume is w*l*h

The surface area (SA) of a box without a top is wl+2wh+2lh.

Since the bottom is square we have l = w so

SA = w^2 + 4wh

And we are told this is 58 cm^2 so

w^2 + 4wh = 58. (1)

The volume of the box

V = w^2 * h (2)

and we can solve (1) for h = (58-w^2)/4w = 29/(2w) - w/4

Substituting into (2) we get

V = w^2(29/(2w) - w/4) = 29/2 w - w^3/4

Taking the derivative

dV/dw = 29/2 - 3/4 w^2

We can have the extrema when dV/dw = 0 so

29/2 - 3/4 w^2 = 0

3/4 w^2 = 29/2

w^2 = 58/3

w = +/- sqrt(58/3)

Since we are talking about a width the negative solution does not make sense, so we get

w = sqrt(58/3) = 4.3969687 cm

h = 29/2sqrt(58/3) - sqrtr(58/3)/4 = 29/2 * sqrt(58/3)/(58/3) - sqrt(58/3)/4

= 3/4 sqrt(58/3) - sqrt(58/3)/4 = 1/2 sqrt(58/3) = 2.1984843 cm

**So the answer is w = sqrt(58/3), h = 1/2 sqrt(58/3)**