Question 111234
Since you are cutting x out of each corner, the length of the box must be 10 - 2x, and the width of the box must be 5 - 2x, and the height is x.  So the volume must be given by:
:
{{{V=x(10-2x)(5-2x)}}}
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The domain of this function is determined by the physical limitations.  In the first place, if x = 0, then there is no box, so one condition on the domain is that x > 0.  On the other hand, if x = 2.5, there also is no box because the width (5 - 2x) would be zero, so the other condition on the domain is x < 2.5.  So the entire expression for the domain is {{{0<x<2.5}}}.
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For the range, on one end we know that the box must have some volume, so x > 0.  On the other end, the maximum value of the range is equal to the maximum Volume.  Therefore, we have to solve the third part of the problem now.
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The volume function can be expressed as:
{{{V=4x^3-30x^2+50x}}}
This function will have a local maximum at the point in the range where the first derivative is zero.
:
{{{dV/dx=12x^2-60x+50}}}
{{{12x^2-60x+50=0}}}
{{{6x^2-30x=-25}}}
{{{x^2-5x=-25/6}}}
{{{x^2-5x+25/4=-25/6+25/4}}}
{{{(x-5/2)^2=25/12}}}
{{{x=5/2+-sqrt(25/12)}}}
{{{x=5/2+(5*sqrt(3)/6)}}}, which is > 2.5, so invalid
{{{x=5/2-(5*sqrt(3)/6)}}}, is the x coordinate of the local maximum
:
{{{V[max]=4(5/2-(5*sqrt(3)/6))^3-30(5/2-(5*sqrt(3)/6))^2+50(5/2-(5*sqrt(3)/6))}}}
:
I'll let you simplify that mess, but it is the upper value of the range, and the answer to the third part of the problem.
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Surface Area Problem:
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The surface area of the flat piece of cardboard is 5 X 10 or 50 sq inches.  But you are cutting out 4 pieces each x * x, so the box surface area is {{{S=50-4x^2}}}.
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Again the domain must be {{{0<x<2.5}}} because if x is anywhere outside of that interval, there is no box.
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The top end of the range is x < 50, and the bottom end is 25 which is the limit of S as x approaches 2.5.