To find the maximum volume of the box, we need to find the value of x that maximizes the function V(x) = x(10 - 2x)(12 - 2x).
To do this, we can take the derivative of V(x) with respect to x and set it equal to 0 to find the critical points. Then, we can use the second derivative test to determine if these critical points correspond to a maximum or minimum.
V(x) = x(10 - 2x)(12 - 2x)
= 4x^3 - 44x^2 + 120x
Taking the derivative:
V'(x) = 12x^2 - 88x + 120
Setting V'(x) = 0:
12x^2 - 88x + 120 = 0
Factoring:
4x^2 - 22x + 30 = 0
(2x - 5)(2x - 6) = 0
Solving for x:
2x - 5 = 0
x = 2.5
2x - 6 = 0
x = 3
Now, we can use the second derivative test to determine if these critical points correspond to a maximum or minimum.
Taking the second derivative:
V''(x) = 24x - 88
For x = 2.5:
V''(2.5) = 24(2.5) - 88
= 60 - 88
= -28
For x = 3:
V''(3) = 24(3) - 88
= 72 - 88
= -16
Since V''(x) < 0 for both critical points, this indicates that the function V(x) has a local maximum at x = 2.5 and x = 3.
To find the maximum volume, we can substitute these critical points into the function V(x).
For x = 2.5:
V(2.5) = 2.5(10 - 2(2.5))(12 - 2(2.5))
= 2.5(10 - 5)(12 - 5)
= 2.5(5)(7)
= 87.5
For x = 3:
V(3) = 3(10 - 2(3))(12 - 2(3))
= 3(10 - 6)(12 - 6)
= 3(4)(6)
= 72
The maximum volume of a box that can be created from the piece of paper is 87.5 cubic inches.
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.
1 answer
1 answer