have 50 sq ft of material to make an open top box with a square base.
a) use formula for surface area to express the height h of the box in terms of x.
b) find the dimensions of the box that will produce the maximum volume.
side of square = x
height = h
surface area = x^2 + 4xh
x^2 + 4xh = 50
so,
h = (50 - x^2)/4x
v = hx^2 = (50-x^2)/4x * x^2
v = x(50-x^2)/4
v has a max at x = 5sqrt(2/3) = 4.08
so, a box 4 x 4 x 17/8 will have maximum volume of 34
To answer these questions, we need to consider the surface area and volume of the open top box with a square base.
a) To find the formula for the surface area, we need to break down the box's dimensions. Let's denote the side length of the square base as x, and the height of the box as h.
The surface area of a box is given by the sum of the areas of its individual faces. For the open top box, we have one square base and four sides (rectangles) forming its walls. The area of the square base is x^2, and the area of each of the four sides is xh.
So, the total surface area A(x,h) can be expressed as:
A(x, h) = x^2 + 4xh
We know that the total surface area is equal to 50 square feet, so we can set up the equation:
50 = x^2 + 4xh
Now, we can rearrange this equation to express the height h in terms of x:
4xh = 50 - x^2
h = (50 - x^2) / (4x)
b) To find the dimensions of the box that will produce the maximum volume, we need to consider the volume formula. The volume of a box is given by the product of its base area and height. In this case, the base area is x^2.
So, the volume V(x) can be expressed as:
V(x) = x^2 * h
Using the expression for h from part a), we can substitute it into the volume equation:
V(x) = x^2 * ((50 - x^2) / (4x))
Simplifying this expression would give us:
V(x) = (50x - x^3) / 4
To find the dimensions of the box that will produce the maximum volume, we need to find the value of x that maximizes the function V(x). To do this, we can take the derivative of V(x) with respect to x, set it equal to zero, and solve for x.