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)
cubic inches
To find the maximum volume, we need to find the maximum value of the function V(x) = x(10-2x)(12-2x).
Taking the first derivative of V(x) with respect to x, we get:
V'(x) = (10-2x)(12-2x) + x(-2)(12-2x) + x(10-2x)(-2)
Simplifying, we get:
V'(x) = (10-2x)(12-2x) - 2x(12-2x) - 2(10-2x)(12-2x)
Expanding further, we have:
V'(x) = 120 - 24x - 20x + 4x^2 - 24x + 4x^2 - 60 + 12x + 60 - 4x^2
Combining like terms, we get:
V'(x) = -28x + 120
To find the critical points of the function, we set V'(x) equal to zero and solve for x:
-28x + 120 = 0
-28x = -120
x = 120/28
x ≈ 4.3
We also know that the maximum volume will occur either at the critical point or at the endpoints of the interval [0, 5] (since we can't have negative dimensions for the box).
To check the endpoints, we evaluate V(x) at x = 0 and x = 5:
V(0) = 0(10-2*0)(12-2*0) = 0
V(5) = 5(10-2*5)(12-2*5) = 5(10-10)(12-10) = 0
Both endpoints yield a volume of 0, so the maximum volume must occur at the critical point x ≈ 4.3.
To find the maximum volume, we substitute x ≈ 4.3 back into the original function:
V(x) = x(10-2x)(12-2x)
V(4.3) ≈ 4.3(10-2*4.3)(12-2*4.3) ≈ 4.3(10-8.6)(12-8.6) ≈ 4.3(1.4)(3.4) ≈ 20.3 cubic inches
Therefore, the maximum volume of the box that can be created from the piece of paper is approximately 20.3 cubic inches, rounded to one decimal place.
To find the maximum volume of a box, we need to maximize the function V(x) = x(10 - 2x)(12 - 2x).
1. Start by expanding the expression: V(x) = x(10 - 2x)(12 - 2x)
= x(120 - 20x - 24x + 4x^2)
= x(4x^2 - 44x + 120)
= 4x^3 - 44x^2 + 120x
2. To find the maximum volume, we need to find the critical points of the function. To do this, take the derivative of V(x) with respect to x:
V'(x) = 12x^2 - 88x + 120
3. Set V'(x) equal to zero and solve for x:
12x^2 - 88x + 120 = 0
4. This quadratic equation can be factored:
4(3x^2 - 22x + 30) = 0
3x^2 - 22x + 30 = 0
(3x - 10)(x - 3) = 0
5. Set each factor equal to zero and solve for x:
3x - 10 = 0
x = 10/3
x - 3 = 0
x = 3
6. The critical points are x = 10/3 and x = 3. To determine which is the maximum, we can use the second derivative test. Take the second derivative of V(x):
V''(x) = 24x - 88
7. Plug in the value of x into V''(x) to determine the concavity of the graph:
V''(10/3) = 24(10/3) - 88 = -8/3 < 0
V''(3) = 24(3) - 88 = -16 < 0
8. Since both values of x give a negative second derivative, the function V(x) is concave down at both points.
9. Therefore, the maximum volume occurs at the critical point x = 3.
10. Substitute x = 3 into the original function V(x) to find the maximum volume:
V(3) = 3(10 - 2(3))(12 - 2(3))
= 3(10 - 6)(12 - 6)
= 3(4)(6)
= 72 cubic inches
Thus, the maximum volume of the box that can be created from the piece of paper is 72 cubic inches.
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).
First, let's simplify the function by expanding it:
V(x) = x(10-2x)(12-2x)
= x(120 - 44x + 4x^2)
= 120x - 44x^2 + 4x^3
Next, we want to find the critical points of the function, which occur when the derivative of the function is equal to zero.
Let's find the derivative of V(x):
V'(x) = 120 - 88x + 12x^2
Setting V'(x) equal to zero:
120 - 88x + 12x^2 = 0
Rearranging the equation:
12x^2 - 88x + 120 = 0
Now we can solve this quadratic equation for x using any method such as factoring, completing the square, or using the quadratic formula.
In this case, let's use factoring:
12x^2 - 88x + 120 = 0
3x^2 - 22x + 30 = 0
(3x - 10)(x - 3) = 0
Setting each factor equal to zero:
3x - 10 = 0 or x - 3 = 0
3x = 10 x = 3
x = 10/3
We now have two potential critical points: x = 10/3 and x = 3.
To find the maximum volume, we need to evaluate the function V(x) at these critical points and at the endpoints of the domain, which is x = 0 and x = 6 (since if x > 6, the box cannot be formed).
V(0) = 0(10-2*0)(12-2*0) = 0
V(6) = 6(10-2*6)(12-2*6) = 0 (since 6 is beyond the domain)
V(3) = 3(10-2*3)(12-2*3) = 3(10-6)(12-6) = 3(4)(6) = 72
V(10/3) = (10/3)(10-2*(10/3))(12-2*(10/3)) = (10/3)(10-20/3)(12-20/3) = (10/3)(10/3)(6/3) = 200/9
Comparing the values, we find that the maximum volume of the box that can be created from the piece of paper is 72 cubic inches.
Therefore, the rounded answer to one decimal place is 72.0 cubic inches.