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.
I found the answer to be height/diameter=1/2 when I searched online but there was no work shown. I need someone to show me works. thanks!
see
http://www.jiskha.com/display.cgi?id=1395265529
vol = pi r^2 h
so
h = V/(pi r^2)
area = pi r^2 + 2 pi r h
area = A = pi r^2 + 2 pi r V/(pi r^2)
A = pi r^2 + 2 V /r
dA/dr = 0 = 2 pi r -2 V /r^2
V / r^2 = pi r
V = pi r^3
but
V = h (pi r^2)
so
h (pi r^2) = pi r^3
h = r
or
h = D/2
LOL - I do not even remember doing that !
To solve this problem, let's break it down into steps:
a) Finding the equation to be maximized or minimized:
Let's consider the volume V and amount of material used M as the functions we want to maximize or minimize, respectively. The volume of a right circular cylinder is given by V = πr^2h, where r is the radius and h is the height. Since we want to find the ratio of the height to the diameter, we know that the diameter is 2r.
To find the amount of material used M, we need to consider both the circular bottom and the cylindrical side of the cup. The area of the circular bottom is given by A1 = πr^2, and the area of the cylindrical side is given by A2 = 2πrh.
To minimize M for a fixed volume, we need to find the ratio of h to r which minimizes M. So, the equation to be minimized is M = A1 + A2 = πr^2 + 2πrh.
b) Finding the solution:
To minimize M, we need to find the critical points by taking the partial derivatives of M with respect to r and h and setting them equal to zero.
∂M/∂r = 2πr + 2πh = 0 (equation 1)
∂M/∂h = 2πr = 0 (equation 2)
From equation 2, we see that r = 0, which doesn't make sense in the context of the problem. So, we'll ignore this solution.
From equation 1, we have r + h = 0, which implies h = -r.
Since h cannot be negative, this solution is also ignored.
Now, let's consider the boundary conditions for the given problem. The volume equation becomes V = πr^2(-r) = -πr^3, which implies that r < 0. However, a negative radius is not physically feasible for this problem, so we disregard the boundary condition.
Since we have disregarded all the possible solutions, we now need to consider the endpoints of the problem. For a fixed volume, r > 0 and h > 0. Taking the limit as r approaches 0 or infinity, we find that M approaches infinity as well. Hence, M is not minimized at the endpoints.
c) Showing that your solution is an absolute max or min:
As we've disregarded all other solutions and found that M is not minimized at the endpoints, the only remaining possibility is that there is no minimum value for M. Therefore, the material used M is minimized when the ratio of the height to the diameter is h/d = 1/2.
Note: It's also worth mentioning that this mathematical result makes physical intuitive sense. By making the height equal to half the diameter, we are using the least amount of material to construct the drinking cup, as more material would be required to increase either the height or the diameter.