To find the maximum volume, we need to find the maximum value of the function V(x)=x(10-2x)(12-2x).
First, let's simplify the function:
V(x) = x(10-2x)(12-2x) = 4x^3 - 44x^2 + 120x.
Now, let's find the critical points by taking the derivative of V(x) and setting it equal to zero:
V'(x) = 12x^2 - 88x + 120 = 0.
We can solve this quadratic equation by factoring:
12x^2 - 88x + 120 = 0
Divide both sides by 4:
3x^2 - 22x + 30 = 0
(3x - 10)(x - 3) = 0
Setting each factor equal to zero gives us two possible critical points:
3x - 10 = 0 => x = 10/3
x - 3 = 0 => x = 3
To determine which critical point is the maximum, we can use the second derivative test. Taking the derivative of V'(x), we get:
V''(x) = 24x - 88.
Plugging in the value of x = 10/3 gives us:
V''(10/3) = 24(10/3) - 88 = 24/3 - 88 = -88/3.
Since the second derivative is negative at x = 10/3, this means that x = 10/3 is a maximum. Therefore, the maximum volume of the box is obtained when x = 10/3.
Plugging this value back into the original function, we get:
V(10/3) = (10/3)(10 - 2(10/3))(12 - 2(10/3))
= (10/3)(10 - 20/3)(12 - 20/3)
= (10/3)(30/3 - 20/3)(36/3 - 20/3)
= (10/3)(10/3)(16/3)
= 1600/81.
Rounding to one decimal place, the maximum volume of the box is approximately 19.8 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 point).
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