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.
___cubic inches
To find the maximum volume, we need to find the critical points of the function V(x) and then determine which one is the maximum.
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)
Simplifying this expression:
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 the derivative equal to zero and solve for x:
(10−2x)(12−2x) - 44x + 8x^2 = 0
Expanding and rearranging the equation:
120 - 24x - 20x + 4x^2 - 44x + 8x^2 = 0
12x^2 - 88x + 120 = 0
3x^2 - 22x + 30 = 0
Now we can solve this quadratic equation for x. Using the quadratic formula, we have:
x = (-(-22) ± sqrt((-22)^2 - 4(3)(30))) / (2(3))
x = (22 ± sqrt(484 - 360)) / 6
x = (22 ± sqrt(124)) / 6
x = (22 ± 2sqrt(31)) / 6
x = (11 ± sqrt(31)) / 3
The two possible values for x are (11 + sqrt(31)) / 3 and (11 - sqrt(31)) / 3.
To determine which value of x gives the maximum volume, we need to check the second derivative of V(x) at these points.
The second derivative of V(x) is given by:
V''(x) = -4(12−2x) - 2(12−2x) + 16x
Let's evaluate V''((11 + sqrt(31)) / 3):
V''((11 + sqrt(31)) / 3) = -4(12−2((11 + sqrt(31)) / 3)) - 2(12−2((11 + sqrt(31)) / 3)) + 16((11 + sqrt(31)) / 3)
V''((11 + sqrt(31)) / 3) = -4(12−22/3 + 2sqrt(31)/3) - 2(12−22/3 + 2sqrt(31)/3) + 16((11 + sqrt(31)) / 3)
V''((11 + sqrt(31)) / 3) = -4(34/3 - 2sqrt(31)/3) - 2(34/3 - 2sqrt(31)/3) + 16((11 + sqrt(31)) / 3)
V''((11 + sqrt(31)) / 3) = -4(34 - 2sqrt(31)) / 3 - 2(34 - 2sqrt(31)) / 3 + 16((11 + sqrt(31)) / 3)
V''((11 + sqrt(31)) / 3) = (-4(34 - 2sqrt(31)) - 2(34 - 2sqrt(31)) + 16(11 + sqrt(31))) / 3
V''((11 + sqrt(31)) / 3) = (68sqrt(31) + 364) / 3
Since the second derivative is positive for ((11 + sqrt(31)) / 3), it means that this point corresponds to a minimum, not a maximum.
Now let's evaluate V''((11 - sqrt(31)) / 3):
V''((11 - sqrt(31)) / 3) = (68sqrt(31) - 364) / 3
Since the second derivative is negative for ((11 - sqrt(31)) / 3), it means that this point corresponds to a maximum.
Therefore, the maximum volume of the box that can be created from the piece of paper is given when x = (11 - sqrt(31)) / 3.
Now we can substitute this value of x back into the original function V(x) to find the maximum volume:
V((11 - sqrt(31)) / 3) = ((11 - sqrt(31)) / 3)(10−2((11 - sqrt(31)) / 3))(12−2((11 - sqrt(31)) / 3))
Calculating this expression, you'll find that the maximum volume is approximately 131.8 cubic inches (rounded to one decimal place).
To find the maximum volume of the box, we need to find the maximum value of the function V(x) = x(10 - 2x)(12 - 2x).
First, let's expand the equation:
V(x) = x(10 - 2x)(12 - 2x)
= x(120 - 24x - 20x + 4x^2)
= x(120 - 44x + 4x^2)
= 4x^3 - 44x^2 + 120x
Now, to find the maximum value, we need to find the critical points. The critical points occur where the derivative of V(x) is equal to zero.
Let's find the derivative of V(x):
V'(x) = 12x^2 - 88x + 120
Setting V'(x) to zero and solving for x:
12x^2 - 88x + 120 = 0
We can simplify this equation by dividing all the terms by 4:
3x^2 - 22x + 30 = 0
Now we can solve this quadratic equation by factoring or using the quadratic formula. Since it doesn't factor nicely, let's use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a = 3, b = -22, and c = 30.
x = (-(-22) ± sqrt((-22)^2 - 4*3*30)) / (2*3)
= (22 ± sqrt(484 - 360)) / 6
= (22 ± sqrt(124)) / 6
≈ 2.6, 4.4
Now we have two critical points, x ≈ 2.6 and x ≈ 4.4. Since we are dealing with a length, we can only consider positive values. Therefore, we'll only consider x ≈ 4.4.
Now, we need to determine if this critical point corresponds to a maximum or minimum. We can do this by checking the sign of the second derivative.
Let's find the second derivative of V(x):
V''(x) = 24x - 88
Evaluate V''(x) at x ≈ 4.4:
V''(4.4) = 24(4.4) - 88
= 105.6 - 88
= 17.6
Since V''(4.4) > 0, the second derivative is positive, indicating a concave-up function. Therefore, the critical point x ≈ 4.4 corresponds to a minimum.
Now, to find the maximum volume, we need to evaluate the function V(x) at this critical point:
V(4.4) = 4(4.4)^3 - 44(4.4)^2 + 120(4.4)
≈ 183.7
Therefore, the maximum volume of the box that can be created from the piece of paper is approximately 183.7 cubic inches, rounded to one decimal place.