Let us assume the base dimension as xft and height of the box as yft.

Volume of the box `(V) = x*x*y`

`170 = x*x*y`

`y = 170/x^2`

We have one square base and four side faces in the box.

If the cost of construction is A;

`A = x^2*7+4*x*y*9`

`A = 7x^2+36*x*170/x^2`

`A = 7x^2+6120/x`

For cost of construction to be maximum or minimum` (dA)/dx = 0`

`(dA)/dx = 14x-6120/x^2`

When `(dA)/dx = 0` ;

`14x-6120/x^2 = 0`

`x^3 = 6120/14`

`x = 7.56`

If the cost is minimum at x = 7.56 then `((d^2A)/(dx^2))_(x=7.56)>0`

`(d^2A)/(dx^2) = 14+12240/x^3`

Since x>0 then `14+12240/x^3>0` always.

`(d^2A)/(dx^2)>0` always. This means A has a minimum.

So the cost is minimum at x = 7.56

`y = 170/7.56^2 = 2.97`

Dimensions

*base = 7.56ft*

*side = 2.97ft*

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