A square piece of canvas measuring 10 ft. by 10 ft. is to be used to make a pup tent with open ends as shown. How long should the center pole be if the volume covered by the tent is to be maximum?

I guess we are not allowed to cut this canvas up so the tent will be ten feet long and the opening at the end will be an equilateral triangle with slopes(hypotenuse of right triangle bounded by center pole h, slope, and ground g, of 5 ft.

Maximize the area of that triangle
A = .5 g h
dA/dh= .5 g dh/dh + .5 h dg/dh
= .5 g + .5 h dg/dh
= 0 for max or min

g^2 + h^2 = 5^2 = 25
g^2 = 25 - h^2
g = sqrt (25-h^2) = (25-h^2)^.5

dg/dh = .5(25-h^2)^-.5 (-2h)
= - h (25-h^2)^-.5
so
0 = .5(25-h^2)^.5 - .5h^2 (25-h^2)^-.5

(25-h^2)^.5 = h^2 /(25-h^2)
25 - h^2 = h^2
2 h^2 = 25
h^2 = 5^2 /2
h = 5/sqrt 2 = 5 sqrt 2/2 = 3.54 ft

To find the length of the center pole that maximizes the volume covered by the tent, we need to determine the dimensions of the tent's rectangular base. Since we have a square piece of canvas, one side of the base will be equal to the length of the side of the canvas, which is 10 ft.

Let's assume the other side of the rectangular base, which will be along the center pole, is x ft. Now we can determine the dimensions of the base:

Width of the base = x ft
Length of the base = 10 ft

The height of the tent, h, will be the length of the center pole. Considering the shape of the tent, the height will be perpendicular to the rectangular base and will be equal to x ft.

Now we have all the dimensions needed to calculate the volume of the tent, V.

V = Length x Width x Height
V = 10 ft x x ft x x ft
V = 10x^2 ft^3

To maximize the volume, we need to maximize the expression 10x^2. The maximum value of this expression will occur when x is at its maximum value. However, we need to consider the constraint that the total length of the canvas is 10 ft.

Since one side of the rectangular base is x ft, and the other side is 10 ft, the total length of the canvas will be the sum of all sides of the rectangle:

Total length of the canvas = x ft + 10 ft + x ft
Total length of the canvas = 2x + 10 ft

So, the length of the center pole should be equal to the total length of the canvas:

Center pole length = 2x + 10 ft

To maximize the volume, we can find the maximum value of 10x^2 by taking the derivative and setting it to zero:

dV/dx = 20x

Setting it equal to zero:

20x = 0
x = 0

Since x represents a length, it cannot be equal to zero, so we need to consider the critical points nearby.

To do this, we can find the second derivative:

d^2V/dx^2 = 20

Since the second derivative is positive, we can conclude that the critical point at x = 0 corresponds to a minimum rather than a maximum.

However, we also need to consider the constraint that the total length of the canvas is 10 ft.

Total length of the canvas = 2x + 10 ft

Setting it equal to 10 ft:

2x + 10 ft = 10 ft
2x = 0 ft
x = 0 ft

Since x = 0 ft satisfies the constraint, it means that one end of the tent is closed, resulting in no volume.

Therefore, to maximize the volume covered by the tent, the length of the center pole should be equal to the total length of the canvas:
Center pole length = 2x + 10 ft

By substituting the value of x into the equation, we find:
Center pole length = 2(10 ft) + 10 ft
Center pole length = 20 ft + 10 ft
Center pole length = 30 ft

Hence, the length of the center pole should be 30 ft to maximize the volume covered by the tent.