# Answer to Question #4842 in Calculus for Liliana Chavez

Question #4842

A box with an open top is to be constructed from a square piece of cardboard, 10 in wide, by cutting out a square from each of the four corners and bending up the sides.

A. What is the length of the side of the square cut out from the corners of the box. Let x = the length of the side of the square cut out of the corners

B. What is the maximum volume of such a box?

A. What is the length of the side of the square cut out from the corners of the box. Let x = the length of the side of the square cut out of the corners

B. What is the maximum volume of such a box?

Expert's answer

The volume of a box is submitted by function V = S*H, where S is an area of the bottom egde and H is a height of the box. The height H is equal to the length of the side of the square cut out of the corners, so H=x. The length of a side of the bottom egde is l = 10-2x, so S = (10-2x)^2. At last, V(x) = x(10-2x)^2 = 4x^3-40x^2+100x.

Let's find the derivative of volume with respect to x: V' = 12x^2-80x+100. Let's consider equation 12x^2-80x+100 = 0. It has two solutions: x=5, which turns V(x) to zero and x=5/3, which turns V(x) to maximum. So, maximum volume of the box is V(5/3) = 10000/135.

Let's find the derivative of volume with respect to x: V' = 12x^2-80x+100. Let's consider equation 12x^2-80x+100 = 0. It has two solutions: x=5, which turns V(x) to zero and x=5/3, which turns V(x) to maximum. So, maximum volume of the box is V(5/3) = 10000/135.

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