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.

Question

___cubic inches

Answer 1

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

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)

Simplifying this expression:
V'(x) = (10−2x)(12−2x) - 2x(12−2x) - 2x(10−2x)
V'(x) = (10−2x)(12−2x) - 24x + 4x^2 - 20x + 4x^2
V'(x) = (10−2x)(12−2x) - 44x + 8x^2

Now, let's set the derivative equal to zero and solve for x:
(10−2x)(12−2x) - 44x + 8x^2 = 0

Expanding and rearranging the equation:
120 - 24x - 20x + 4x^2 - 44x + 8x^2 = 0
12x^2 - 88x + 120 = 0
3x^2 - 22x + 30 = 0

Now we can solve this quadratic equation for x. Using the quadratic formula, we have:
x = (-(-22) ± sqrt((-22)^2 - 4(3)(30))) / (2(3))
x = (22 ± sqrt(484 - 360)) / 6
x = (22 ± sqrt(124)) / 6
x = (22 ± 2sqrt(31)) / 6
x = (11 ± sqrt(31)) / 3

The two possible values for x are (11 + sqrt(31)) / 3 and (11 - sqrt(31)) / 3.

To determine which value of x gives the maximum volume, we need to check the second derivative of V(x) at these points.

The second derivative of V(x) is given by:
V''(x) = -4(12−2x) - 2(12−2x) + 16x

Let's evaluate V''((11 + sqrt(31)) / 3):
V''((11 + sqrt(31)) / 3) = -4(12−2((11 + sqrt(31)) / 3)) - 2(12−2((11 + sqrt(31)) / 3)) + 16((11 + sqrt(31)) / 3)
V''((11 + sqrt(31)) / 3) = -4(12−22/3 + 2sqrt(31)/3) - 2(12−22/3 + 2sqrt(31)/3) + 16((11 + sqrt(31)) / 3)
V''((11 + sqrt(31)) / 3) = -4(34/3 - 2sqrt(31)/3) - 2(34/3 - 2sqrt(31)/3) + 16((11 + sqrt(31)) / 3)
V''((11 + sqrt(31)) / 3) = -4(34 - 2sqrt(31)) / 3 - 2(34 - 2sqrt(31)) / 3 + 16((11 + sqrt(31)) / 3)
V''((11 + sqrt(31)) / 3) = (-4(34 - 2sqrt(31)) - 2(34 - 2sqrt(31)) + 16(11 + sqrt(31))) / 3
V''((11 + sqrt(31)) / 3) = (68sqrt(31) + 364) / 3

Since the second derivative is positive for ((11 + sqrt(31)) / 3), it means that this point corresponds to a minimum, not a maximum.

Now let's evaluate V''((11 - sqrt(31)) / 3):
V''((11 - sqrt(31)) / 3) = (68sqrt(31) - 364) / 3

Since the second derivative is negative for ((11 - sqrt(31)) / 3), it means that this point corresponds to a maximum.

Therefore, the maximum volume of the box that can be created from the piece of paper is given when x = (11 - sqrt(31)) / 3.

Now we can substitute this value of x back into the original function V(x) to find the maximum volume:

V((11 - sqrt(31)) / 3) = ((11 - sqrt(31)) / 3)(10−2((11 - sqrt(31)) / 3))(12−2((11 - sqrt(31)) / 3))

Calculating this expression, you'll find that the maximum volume is approximately 131.8 cubic inches (rounded to one decimal place).