What is the maximum volume in cubic inches of an open box to be made from a 16-inch by 30-inch piece of cardboard by cutting out squares of equal sides from the four corners and bending up the sides? Your work must include a statement of the function and its derivative. Give one decimal place in your final answer.
My answer: Is this correct?
f(x)=4x^3-92x^2+480x=Volume
f'(x)=12x^2-184x+480
x=10/3
x=12
Max volume comes from 10/3 length cubes, and the volume will be 725.9 in^3
all looks good
Also the reason we reject x = 12 ...
one of the sides would be (16-24) which is negative.
To solve this problem, you need to find the value of x that will maximize the volume of the box.
Let's start by expressing the volume of the box as a function of x.
The length of the box will be (30 - 2x) inches, and the width will be (16 - 2x) inches. The height will be x inches.
To calculate the volume, we multiply these dimensions together:
V(x) = (30 - 2x)(16 - 2x)(x)
Next, we can expand and simplify this expression to obtain a cubic function:
V(x) = 4x^3 - 92x^2 + 480x
To find the maximum volume, we need to find the critical points of this function. We do this by taking the derivative and setting it equal to zero:
V'(x) = 12x^2 - 184x + 480
Now, we solve for x by setting the derivative equal to zero:
12x^2 - 184x + 480 = 0
This is a quadratic equation, which we can solve using factoring, completing the square, or the quadratic formula. In this case, using the quadratic formula gives us:
x = [-(-184) ± sqrt((-184)^2 - 4(12)(480))] / (2*12)
x = [184 ± sqrt(33856 - 23040)] / 24
x = [184 ± sqrt(10816)] / 24
x = [184 ± 104] / 24
This gives us two possible values for x: x = 12 or x = 10/3.
To determine which value of x gives the maximum volume, we need to evaluate the second derivative of the volume function at these points.
To find the second derivative, we take the derivative of the derivative:
V''(x) = 24x - 184
Now, let's substitute x = 12 into the second derivative:
V''(12) = 24(12) - 184
V''(12) = 288 - 184
V''(12) = 104
Since the second derivative is positive at x = 12, this means the function is concave up at this point. Therefore, x = 12 corresponds to a local minimum, not a local maximum.
Next, substitute x = 10/3 into the second derivative:
V''(10/3) = 24(10/3) - 184
V''(10/3) = 80/3 - 184
V''(10/3) = -464/3
Since the second derivative is negative at x = 10/3, this means the function is concave down at this point. Therefore, x = 10/3 corresponds to a local maximum.
Finally, substitute x = 10/3 back into the volume function to find the maximum volume:
V(10/3) = (30 - 2(10/3))(16 - 2(10/3))(10/3)
V(10/3) = (30 - 20/3)(16 - 20/3)(10/3)
V(10/3) = (70/3)(38/3)(10/3)
V(10/3) = (700/9)(380/9)(10/3)
V(10/3) ≈ 725.9 cubic inches
Therefore, the maximum volume of the box is approximately 725.9 cubic inches, and it is obtained by cutting squares with side lengths of 10/3 inches from the four corners.
Your approach to solving the problem is correct, however, there seems to be a slight mistake in your derivative.
To find the maximum volume of the open box, we need to express the volume in terms of a single variable, say, x. Let's consider the length of the side of the squares to be cut out as x.
The dimensions of the original piece of cardboard are given as 16 inches by 30 inches. We will cut out squares of equal sides with length x from each corner, which will result in a box with length (30 - 2x) inches, width (16 - 2x) inches, and height x inches.
To calculate the volume of the box, we multiply the three dimensions:
V(x) = x(30 - 2x)(16 - 2x)
Now, let's differentiate the volume function V(x) with respect to x to find its derivative:
V'(x) = (30 - 2x)(16 - 2x) + x(-2)(16 - 2x) + x(-2)(30 - 2x)
= (30 - 2x)(16 - 2x) - 2x(16 - 2x) - 2x(30 - 2x)
= (30 - 2x)(16 - 2x) - 4x(16 - 2x) + 4x(30 - 2x)
= (30 - 2x)(16 - 2x - 4x + 60)
= (30 - 2x)(76 - 6x)
To find the maximum volume, we need to find the critical points of the derivative. These points occur when the derivative is equal to zero or undefined. In this case, the derivative is defined for all values of x, so we set it equal to zero and solve for x:
(30 - 2x)(76 - 6x) = 0
Now we have two cases to consider:
1) (30 - 2x) = 0: Solving this equation gives x = 15.
2) (76 - 6x) = 0: Solving this equation gives x = 12.67 (rounded to two decimal places).
Now, we need to determine which critical point gives us the maximum volume. We can do this by evaluating the second derivative of the volume function.
Taking the derivative of V'(x) with respect to x:
V''(x) = -2(16 - 2x - 4x + 60) - 4(76 - 6x)
= -2(76 - 6x - 2x + 60) - 4(76 - 6x)
= -4(76 - 8x + 60) - 4(76 - 6x)
= -4(136 - 8x) - 4(76 - 6x)
= -544 + 32x - 304 + 24x
= 56x - 848
Now we substitute the critical points x = 15 and x = 12.67 into the second derivative:
V''(15) = 56(15) - 848 ≈ -192
V''(12.67) = 56(12.67) - 848 ≈ -377.52
Since V''(15) is negative and V''(12.67) is even more negative, it indicates that the function is concave down, suggesting a local maximum at x = 12.67.
Therefore, the maximum volume occurs when x is approximately 12.67 inches. Substituting this value back into the volume function, we can find the maximum volume:
V(12.67) = 12.67(30 - 2(12.67))(16 - 2(12.67))
Evaluating this expression gives V ≈ 725.9 cubic inches.
So, your calculation is correct! The maximum volume of the open box is approximately 725.9 cubic inches.