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 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.
To find the maximum volume of the box, we need to find the critical points of the function and determine which one yields the maximum value.
1. Start by finding the derivative of the function:
V(x) = x(10 - 2x)(12 - 2x)
To simplify the calculation, we can expand the equation first:
V(x) = x(120 - 20x - 24x + 4x^2)
V(x) = x(4x^2 - 44x + 120)
V(x) = 4x^3 - 44x^2 + 120x
Now, differentiate using the power rule of differentiation:
V'(x) = 12x^2 - 88x + 120
2. Set the derivative equal to zero to find the critical points:
12x^2 - 88x + 120 = 0
We can simplify this quadratic equation by dividing through by 4:
3x^2 - 22x + 30 = 0
3. Solve the quadratic equation for x. You can either factor or use the quadratic formula. In this case, factoring is easier:
(3x - 10)(x - 3) = 0
Set each factor equal to zero:
3x - 10 = 0 or x - 3 = 0
Solve for x:
3x = 10 or x = 3
x = 10/3
We have two critical points: x = 10/3 and x = 3.
4. To determine which critical point gives the maximum volume, we can evaluate the second derivative at each point:
V''(x) = 24x - 88
Evaluate V''(x) at each critical point:
V''(10/3) = 24(10/3) - 88
V''(3) = 24(3) - 88
Simplify:
V''(10/3) = 80/3 - 88
V''(3) = 72 - 88
Simplify further:
V''(10/3) = -8/3
V''(3) = -16
5. Since we are looking for the maximum volume, we need the second derivative to be negative. Only if the second derivative is negative at a critical point, it corresponds to the maximum value.
Given that V''(10/3) = -8/3, this critical point does produce a maximum volume.
6. Plug this critical point back into the original equation to find the maximum volume:
V(10/3) = (10/3)(10 - 2(10/3))(12 - 2(10/3))
Simplify the equation:
V(10/3) = (10/3)(10 - 20/3)(12 - 20/3)
V(10/3) = (10/3)(30/3 - 20/3)(36/3 - 20/3)
V(10/3) = (10/3)(10/3)(16/3)
Multiply the fractions:
V(10/3) = (160/9) cubic inches
7. Round the answer to one decimal place:
V(10/3) ≈ 17.8 cubic inches
Therefore, the maximum volume of the box that can be created from the piece of paper is approximately 17.8 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).
One way to find the maximum is by taking the derivative of the function and setting it equal to zero. The derivative of V(x) is the rate of change of V(x) with respect to x.
Let's find the derivative of V(x):
V'(x) = (10 - 2x)(12 - 2x) + x(-2)(12 - 2x) + x(10 - 2x)(-2)
Simplifying:
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 V'(x) equal to zero and solve for x:
0 = (10 - 2x)(12 - 2x) - 44x + 8x^2
0 = 120 - 24x - 20x + 4x^2 - 44x + 8x^2
0 = 12x^2 - 88x + 120
Now, we can solve this quadratic equation to find the values of x:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 12, b = -88, and c = 120.
x = (-(-88) ± √((-88)^2 - 4(12)(120))) / (2(12))
x = (88 ± √(7744 - 5760)) / 24
x = (88 ± √1984) / 24
Now, we need to find the value of x that gives us the maximum volume. Since we are dealing with a physical object, we discard the negative value of x as it doesn't make sense in this context.
Let's calculate the value of x using the positive square root:
x = (88 + √1984) / 24
x ≈ 3.2 (rounded to one decimal place)
Substituting this value of x back into the original function V(x):
V(3.2) = 3.2(10 - 2(3.2))(12 - 2(3.2))
V(3.2) ≈ 308.8
Therefore, the maximum volume of the box that can be created from the piece of paper is approximately 308.8 cubic inches.