a) To find the equation that represents the amount of material used, we need to consider both the circular bottom and the cylindrical side of the cup.
Let's assume that the cup has a volume V, radius r, height h, and diameter d.
The volume of a right circular cylinder is given by V = πr^2h. Since the volume is fixed, we can express the height in terms of the radius as h = V / (πr^2).
Now, we need to find the equation that represents the amount of material used.
For the circular bottom, the area is given by A_bottom = πr^2.
For the cylindrical side, the area is given by A_side = 2πrh.
The total amount of material used, A_total, is the sum of A_bottom and A_side:
A_total = A_bottom + A_side
= πr^2 + 2πrh
= πr^2 + 2πr (V / (πr^2))
= πr^2 + 2V/r.
Therefore, the equation that represents the amount of material used is A_total = πr^2 + 2V/r.
b) To find the ratio of the height to the diameter that minimizes the amount of material used, we can minimize the equation A_total = πr^2 + 2V/r with respect to r.
To find the minimum, we need to take the derivative of A_total with respect to r and set it equal to zero:
dA_total/dr = 2πr - 2V/r^2 = 0.
Multiplying through by r^2, we get:
2πr^3 - 2V = 0.
Simplifying further, we find:
r^3 = V / π.
Taking the cube root of both sides, we get:
r = (V / π)^(1/3).
Now, using the expression for h in terms of r, we can find the corresponding height:
h = V / (πr^2) = V / (π[(V / π)^(1/3)]^2) = V^(1/3) / π^(2/3).
Finally, we can find the ratio of the height to the diameter by dividing h by 2r:
(height/diameter) = (V^(1/3) / π^(2/3)) / [2(V / π)^(1/3)]
= (V^(1/3) / (2V)^(1/3)) * (π^(1/3) / π^(1/3))
= (V^(1/3) / (2V)^(1/3))
= (1/2)^(1/3)
= 0.7937 (approximately).
Therefore, the ratio of the height to the diameter that minimizes the amount of material used is approximately 0.7937.
c) To show that this solution is an absolute minimum, we need to examine the second derivative of the equation A_total with respect to r.
Taking the second derivative, we have:
d^2A_total/dr^2 = 2π - 4V/r^3.
Substituting the value of r that minimizes the equation A_total, which is r = (V / π)^(1/3), we find:
d^2A_total/dr^2 = 2π - 4V / [(V / π)^(1/3)]^3
= 2π - 4V / (V / π)
= 2π - 4π^2 / V.
Since the volume V is fixed, the value of d^2A_total/dr^2 is positive. Therefore, the ratio of height to diameter that minimizes the amount of material used is an absolute minimum.