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.
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.
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).
Step 1: Expand the equation.
V(x) = x(10 - 20x + 4x^2)(12 - 2x)
Step 2: Simplify the equation.
V(x) = x(120 - 20x - 240x + 40x^2 - 24x + 4x^2)
V(x) = x(40x^2 - 264x + 120)
Step 3: Rearrange the equation to standard form.
V(x) = 40x^3 - 264x^2 + 120x
Step 4: Calculate the derivative of V(x) with respect to x.
V'(x) = 120x^2 - 528x + 120
Step 5: Set V'(x) = 0 and solve for x.
120x^2 - 528x + 120 = 0
Step 6: Solve the quadratic equation either by factoring, completing the square, or using the quadratic formula.
Since the coefficient of x^2 is greater than 1, let's use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values: x = (-(-528) ± √((-528)^2 - 4 * 120 * 120)) / 2 * 120
x = (528 ± √(278784 - 57600)) / 240
x = (528 ± √221184) / 240
x ≈ (528 ± 470.644) / 240
Step 7: Calculate the values of x.
x ≈ (528 + 470.644) / 240 ≈ 3.825
x ≈ (528 - 470.644) / 240 ≈ 0.237
Step 8: Plug the values of x into V(x) to find the corresponding y-values.
V(3.825) ≈ 43.75 cubic inches
V(0.237) ≈ 42.67 cubic inches
Step 9: Compare the two values of V(x). The larger value represents the maximum volume.
Therefore, the maximum volume of the box that can be created from the piece of paper is approximately 43.8 cubic inches (rounded to one decimal place).