SOLUTION: A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 5 in. by 10 in. by cutting out equal squares of side x at each corner and then fol

Algebra ->  Volume -> SOLUTION: A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 5 in. by 10 in. by cutting out equal squares of side x at each corner and then fol      Log On


   

Question 111234: A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 5 in. by 10 in. by cutting out equal squares of side x at each corner and then folding up the sides as in the figure.
(a) Express the volume V of the box as a function of x, then determine the range and the domain of this function.

(b) Express the surface area S of the box as a function of x, then determine the range and the domain of this function.
(c) Find the value of x that gives the largest volume of the box
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
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:
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V=x%2810-2x%29%285-2x%29
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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%3Cx%3C2.5.
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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%5E3-30x%5E2%2B50x
This function will have a local maximum at the point in the range where the first derivative is zero.
:
dV%2Fdx=12x%5E2-60x%2B50
12x%5E2-60x%2B50=0
6x%5E2-30x=-25
x%5E2-5x=-25%2F6
x%5E2-5x%2B25%2F4=-25%2F6%2B25%2F4
%28x-5%2F2%29%5E2=25%2F12
x=5%2F2%2B-sqrt%2825%2F12%29
x=5%2F2%2B%285%2Asqrt%283%29%2F6%29, which is > 2.5, so invalid
x=5%2F2-%285%2Asqrt%283%29%2F6%29, is the x coordinate of the local maximum
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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%5E2.
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Again the domain must be 0%3Cx%3C2.5 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.