The volume of an open top box is created by cutting a square from each corner of a 10 in. by 12 in. piece of paper. The situation is represented by the function V(x)=x(10−2x)(12−2x). Find the maximum volume of a box that can be created from the piece of paper. Round the answer to one decimal place.

Answer 1

To find the maximum volume, we need to find the critical points of the function V(x) and then determine which one corresponds to the maximum volume.

First, let's find the derivative of V(x) with respect to x:
V'(x) = (10 - 2x)(12 - 2x) + x(-2)(12 - 2x) + x(10 - 2x)(-2) = -4x^2 + 44x - 96

Next, we set V'(x) equal to zero and solve for x:
-4x^2 + 44x - 96 = 0
Divide through by -4:
x^2 - 11x + 24 = 0
Factor the quadratic equation:
(x - 3)(x - 8) = 0

So x = 3 or x = 8

We also need to check the endpoints of the interval [0, 5] because we cannot have negative dimensions.

V(0) = 0
V(5) = 5(10 - 2*5)(12 - 2*5) = 5(10 - 10)(12 - 10) = 0

Now we can compare the values of V(3), V(5), and V(8) to determine the maximum volume:

V(3) = 3(10 - 2*3)(12 - 2*3) = 3(10 - 6)(12 - 6) = 3(4)(6) = 72
V(5) = 5(10 - 2*5)(12 - 2*5) = 5(10 - 10)(12 - 10) = 0
V(8) = 8(10 - 2*8)(12 - 2*8) = 8(10 - 16)(12 - 16) = 8(-6)(-4) = 192

So the maximum volume of the box that can be created from the piece of paper is 192 in^3.