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**[SOLVED] Maximum area/volume problems**

I have two here. One I feel I've solved but am just looking for reassurance (I'm needy that way!), The other I've got an answer but I know I've missed something.

## Homework Statement

1. A rectangle is inscribed in a 6/8/10 cm right-angle triangle where two vertices (and one side) touch the hypotenuse side and the other two vertices each touching one of the other two sides.

Find the greatest area posssible.

2.A closed rectangular container with a square base is to have a volume of 2000cm[tex]^{3}[/tex]. It costs twice as much per cm[tex]^{2}[/tex] for the top and bottom as it does for the sides. Find the dimensions of least cost.

## The Attempt at a Solution

For the first one, I made the hypotenuse the base of the triangle. I then worked out the height of the triangle using trig (height = 4.8cm)

I then used the equivalent triangles theory to state that the base: height ratio of the 6/8/10 triangle is 25:12 (10:4.8), therefore the same ratio would apply for the triangle made within by the rectangle, at the top of the large triangle (hope that's clear!).

i.e. If we take the height of the smaller inner triangle to be h and the base b, then the ratio of b:h is 25:12, making b=25h/12. This is one side of the rectangle.

The height of the rectangle is 4.8-h, making the area b*(4.8-h) = 25h/12(4.8-h)

f(h) = 10h - 25h[tex]^{2}[/tex]/12

f'(h) = 10 - 25h/6

putting f'(h) to 0, we get h=2.4, making b=5 and the area 24cm[tex]^{2}[/tex].

Now assuming this is correct, I was curiously wondering if there's any other way of doing this. I justa sorta madea this uppa as I went.

2. base of box = x, height = h.

Volume of box = x[tex]^{2}[/tex]h = 2000, h = 2000/[tex]x^{2}[/tex]

Surface area of box = 2x[tex]^{2}[/tex] + 4xh

SA = 2x[tex]^{2}[/tex] + 4x.2000/[tex]x^{2}[/tex] = 2x[tex]^{2}[/tex] + 8000/x

f'(SA) = 4x - 8000/[tex]x^{2}[/tex] = 0

4[tex]x^{3}[/tex] - 8000 = 0

[tex]x^{3}[/tex] = 2000

x = 12.6, which means h also = 12.6

This is correct in every aspect except it ignores the extra bit of the cost of the box sides/base. Where/how do I bring that info into the equation?