A drinking cup is made in the shape of a right circular cylinder. for a fixed volume, we wish to make the total material used, the circular bottom and the cylindrical side, as small as possible. Find the ratio of the height to the diameter that minimizes the amount of material used. Hint: Express the height and diameter as a function of the radius r and find the value of r that minimizes the amount of material used.
a) find the equation to maximized or minimized.
b) find the solution
c) showing that your solution is an absolute max or min.
v = pi r^2 h, so
h = v/(pi r^2)
the material used is determined by the surface area
a = 2pi r^2 + 2pi rh
= 2pi r^2 + 2pi r(v/(pi r^2))
= 2pi r^2 + 2v/r
minimum area is when da/dr = 0, so we want
4pi r - 2v/r^2 = 0
2pi r^3 - v = 0
r = ∛(v/(2pi))
See the curve (with v=1) at
http://www.wolframalpha.com/input/?i=2pi+r^2+%2B+2pi+r%281%2F%28pi+r^2%29%29
a) To find the equation to maximize or minimize, we need to express the amount of material used as a function of the height and diameter.
Let's assume the radius of the circular base is r, the height is h, and the diameter is d.
The formula for the volume V of a right circular cylinder is V = πr^2h.
The surface area of the bottom of the cup is the area of a circle, A_bottom = πr^2.
The surface area of the side of the cup is the circumference of a circle multiplied by the height, A_side = 2πrh.
The total amount of material used, M, is the sum of the surface areas of the bottom and the side:
M = A_bottom + A_side
M = πr^2 + 2πrh
b) To find the solution, we need to minimize the amount of material used, so we can consider M as a function of either r or h. Let's do it in terms of r.
M(r) = πr^2 + 2πrh
To minimize M(r), we can take the derivative with respect to r and set it equal to zero:
dM/dr = 2πr + 2πh(dr/dr) = 2πr + 2πh(0)
dM/dr = 2πr
Setting this equal to zero to find the critical point:
2πr = 0
r = 0
However, we know that the radius cannot be zero, so this is not a valid solution.
c) To show that our solution is an absolute minimum, we will take the second derivative and evaluate it at the critical point.
d^2M/dr^2 = 2π > 0
Since the second derivative is positive, it means that the function is concave up, which implies that the critical point r = 0 is a minimum.
Therefore, the solution to minimize the amount of material used is when the height to diameter ratio is 0.
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.
a) To find the equation that represents the amount of material used, we need to express the height and diameter in terms of the radius, and then determine the formula for the total surface area of the cup.
Let's assume that the radius of the circular base of the cup is r, the height of the cup is h, and the diameter is d.
The diameter is simply twice the radius: d = 2r.
The total surface area consists of the curved surface area of the cylindrical side of the cup and the area of the circular bottom.
The curved surface area of the cylinder is given by the formula: A_cylinder = 2πrh.
The area of the circular bottom is given by the formula: A_circle = πr^2.
The total surface area of the cup, A_total, is the sum of the curved surface area and the area of the circular bottom: A_total = A_cylinder + A_circle.
b) To find the solution, we need to express the total surface area as a function of one variable, which will be the radius r.
Substituting the formulas for A_cylinder and A_circle into the equation for A_total, we get:
A_total = 2πrh + πr^2.
To minimize the amount of material used, we want to find the minimum value of A_total. We can do this by finding the critical points of A_total with respect to r.
Taking the derivative of A_total with respect to r and setting it equal to zero, we get:
dA_total/dr = 2πh + 2πr = 0.
Simplifying the equation, we find:
r = -h.
However, since we are dealing with physical dimensions, the radius cannot be negative, so we can ignore this solution.
Therefore, we don't have any critical points within the valid range of r.
c) Since we don't have any critical points, we need to consider the boundaries of the valid range of r.
The radius r cannot be negative, and it also cannot be zero, as we need a non-zero cup.
Therefore, we can conclude that there is no absolute minimum or maximum for the ratio of height to diameter that minimizes the amount of material used. The cup's dimensions can vary indefinitely, as long as the ratio of h to d is constant.