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A manufacturer wants to design an open box having a square base and a surface area of 216 square inches. What dimensions will produce a box with maximum volume?

This is what I got when I solved it and is it correct?

V=x^2h

S= (area of base)+ (area of four sides)

S=x^2 + 4xh = 216

V=x^2h

=x^2 (216-x^2/4x)= 54x- x^3/4

0< x < square root 216

dV/dx = 54- 3x^2= 0

then 3x^2 = 216 (when divided it's 72)

x= -9, -8

The lines

V=x^2h

S= (area of base)+ (area of four sides)

S=x^2 + 4xh = 216

V=x^2h

all look OK.

Solving for h gives h = (216 - x^2)/(4x) { note the use of parenthesis is necessary }.

The next line would then be h = 54/x - x/4.

This can be put into V = x^2 h giving V(x) = x^2(54/x - x/4).

That works out to V(x) = 54x - x^3 /4.

This then give dV/dx = 54 - 3x^2 /4.

Solving give 3x^2 = 216.

Yes, this works out at this, but things weren't written correctly in between.

Also, the solution wasn't found correctly.

Now if 3x^2 = 216, then x^2 = 72, so x = ±6*sqrt(2).

That can be put into h = 54/x - x/4 to find h.

Once this has been done, put both x and h into V = 2 x^2 h.

Calculus

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