To find the dimensions of the box that maximizes volume while using 300 square feet of material, we start by defining the variables:
- x: the length of one side of the square base.
- h: the height of the box.
The surface area of the box (which has no top) can be expressed as:
Surface Area = x² + 4xh = 300
Here, x² represents the area of the base, and 4xh is the area of the four sides.
Next, we need to express the volume V of the box:
V = x²h
Now, we will solve for h using the surface area equation:
4xh = 300 – x²
From here, isolate h:
h = (300 – x²) / (4x)
Substituting this value of h into the volume equation, we get:
V = x²((300 – x²) / (4x))
Simplifying this, we can express volume as:
V = (300x – x³) / 4
Now, to find the maximum volume, we take the derivative of V with respect to x, set it to zero, and solve for x:
V’ = (300 – 3x²) / 4 = 0
Solving this gives:
300 – 3x² = 0 -> 3x² = 300 -> x² = 100 -> x = 10
Now substituting x = 10 back into the equation for h:
h = (300 – 10²) / (4 * 10) = (300 – 100) / 40 = 200 / 40 = 5
Thus, the dimensions for maximum volume are:
- Base side length (x): 10 feet
- Height (h): 5 feet
Finally, we can calculate the maximum volume:
V = x²h = 10² * 5 = 100 * 5 = 500
In conclusion, the dimensions of the box that maximizes the volume with a surface area of 300 square feet are a base of 10 feet by 10 feet and a height of 5 feet, yielding a maximum volume of 500 cubic feet.