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).
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.
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 function:
V(x) = x(10-2x)(12-2x)
= x(120 - 24x - 20x + 4x^2)
= x(4x^2 - 44x + 120)
= 4x^3 - 44x^2 + 120x
Step 2: Take the derivative of the function:
V'(x) = 12x^2 - 88x + 120
Step 3: Set the derivative equal to zero and solve for x:
12x^2 - 88x + 120 = 0
Step 4: Factor the quadratic equation:
4(3x^2 - 22x + 30) = 0
3x^2 - 22x + 30 = 0
Step 5: Solve for x using the quadratic formula:
x = (-(-22) ± √((-22)^2 - 4(3)(30))) / (2(3))
x = (22 ± √(484 - 360)) / 6
x = (22 ± √(124)) / 6
x = (22 ± 2√(31)) / 6
x = (11 ± √(31)) / 3
Step 6: Calculate the volume using the value(s) of x:
V(11 + √(31)) = (11 + √(31))(4(11 + √(31))^2 - 44(11 + √(31)) + 120)
V(11 + √(31)) ≈ 209.3
V(11 - √(31)) = (11 - √(31))(4(11 - √(31))^2 - 44(11 - √(31)) + 120)
V(11 - √(31)) ≈ 209.3
The maximum volume of the box that can be created from the piece of paper is approximately 209.3 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).
Step 1: Expand the equation.
V(x) = x(10-2x)(12-2x)
= x(120 - 24x - 20x + 4x^2)
= x(4x^2 - 44x + 120)
= 4x^3 - 44x^2 + 120x
Step 2: Take the derivative of V(x) with respect to x.
V'(x) = 12x^2 - 88x + 120
Step 3: Set V'(x) equal to zero and solve for x.
12x^2 - 88x + 120 = 0
Using the quadratic formula, we can solve for x:
x = (88 ± √(88^2 - 4(12)(120))) / (2(12))
x = (88 ± √(7744 - 5760)) / 24
x = (88 ± √(1984)) / 24
x = (88 ± 44.5) / 24
We have two possible solutions:
x = (88 + 44.5) / 24 = 6.1875
x = (88 - 44.5) / 24 = 1.6042
Step 4: Plug the values of x into the original equation to find the corresponding volumes.
V(6.1875) = 6.1875(10-2(6.1875))(12-2(6.1875))
= 6.1875(10-12.375)(12-12.375)
= 6.1875(-2.375)(-0.375)
= 5.527
V(1.6042) = 1.6042(10-2(1.6042))(12-2(1.6042))
= 1.6042(10-3.2084)(12-3.2084)
= 1.6042(6.7916)(8.7916)
= 90.901
Step 5: Compare the volumes to find the maximum.
The larger volume between 5.527 and 90.901 is 90.901.
Therefore, the maximum volume of the box that can be created from the given piece of paper is approximately 90.9 cubic inches (rounded to one decimal place).