This is a continuation of the question I answered for you earlier.
Since you titled it "Calculus" you should be able to finish it.
1. expand and simplify the expression for V
2. differentiate, you will have a quadratic
3. solve that quadratic by setting it equal to zero for a max of V
4. sub the value you found in 3. into the original volume equation
for the restriction, all you have to do is look at the equation for V
V of course has to be positive.
so x > 0 and x < 3 or else the base side values make no sense.
restriction on x : 0 < x < 3
A manufacturer constructs open boxes from sheets of cardboard that are 6 inches square by cutting small squares from the corners and folding up the sides. The Research and Development Department asks you to determine the size of the square that produces a box of greatest volume. Proceed as follows. Let x be the length of a side of the square to be cut and V be the volume of the resulting box. Show that V = x(6-2x)^2 (sketched it already.)
Are there any restrictions on the value of x? Explain.
Estimate the largest volume.
2 answers
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