Optimization Problem:
There is 300 square feet to construct an open top box with a square base. What dimensions are needed to maximize the volume, and prove that you have found the maximum
x^2 + 4xh = 300
v = x^2 h = x^2 (300-x^2)/4x
dv/dx = 5/4 x^2 (180-x)
max v is where dv/dx = 0, at x=√180
To solve this optimization problem, we need to find the dimensions of the square base that will maximize the volume of the open top box.
Let's define the dimensions as follows:
- Let x be the length of each side of the square base.
- Let h be the height of the box.
First, we need to express the volume of the box in terms of x and h. The volume V of a rectangular box is given by the formula:
V = x^2 * h
We are given that the total available square footage for the open top box is 300 square feet. Since the base and the sides of the box are open, the surface area is equal to the sum of the areas of the four sides:
Surface Area = x^2 + 4xh
Since the total surface area is 300 square feet, we have the equation:
x^2 + 4xh = 300
Now, let's solve the equation for h and substitute it back into the volume formula.
1. Solve the equation for h:
4xh = 300 - x^2
h = (300 - x^2) / 4x
2. Substitute h back into the volume formula:
V = x^2 * [(300 - x^2) / 4x]
V = (1/4)(300x - x^3)
To find the maximum volume, we need to find the critical points of the volume function. We can do this by finding the derivative of the volume function with respect to x and setting it equal to zero.
dV/dx = 0
(1/4)(300 - 3x^2) = 0
300 - 3x^2 = 0
3x^2 = 300
x^2 = 100
x = ±10
Since the dimensions of a physical object cannot be negative, the valid value for x is 10.
Now, substitute this value back into the equation for h to find the corresponding height:
h = (300 - x^2) / 4x
h = (300 - 10^2) / (4 * 10)
h = 200 / 40
h = 5
Therefore, the dimensions that optimize the volume of the open top box are x = 10 and h = 5. To prove that this gives the maximum volume, we can use the second derivative test or substitute other x values into the volume equation and compare the results.