Optimization Problem

A right circular cylindrical can of volume 128tπ cm^3 is to be manufactured by a company to store their newest kind of soup. They want to minimize the surface area of the can to keep costs down.

What are the dimensions of the can with minimum surface area?

The volume of a cyllinder is V= πr^2h, where r is the radius and h is height. The surface area of a cylinder is SA= 2πr^2+2πrh, which is the sum of the area of the top and bottom (2 circles) and the area of the other curved sides (a rectangle, whose length is the circumference of the circles)

They have provided you with all the formulas you need

given: πr^2h = 128
h = 128/(πr^2)

so in
SA = 2πr^2+ 2πrh
= 2πr^2+ 2πr(128/(πr^2)
= 2πr^2 + 256/r
d(SA)/dr = 4πr - 256/r^2
= 0 for a min SA
4πr = 256/r^2
r^3 =64/π
r = 4/π^(1/3) = appr 2.7311
then h = 128/(πr^2) = 5.46223

notice that this is twice the radius
So the minimum SA is obtained when the
radius is 2.4311 cm
and the height is 5.46223 cm

given: πr^2h = 128π

h = 128π/(πr^2)

TSA = 2πr^2+ 2πrh
= 2πr^2+ 2πr(128/(r^2))

d(tsa)/dr = 4πr - (256π/r^2)

0 = 4πr - (256π/r^2)
4πr = (256π/r^2)
(solve for r)
r^3 = 64
r = 4

(solve for h)
h = 128/(4^2)
= 8

To find the dimensions of the can with minimum surface area, we need to optimize the surface area function.

Given:
Volume of the can, V = 128tπ cm^3

Let's start by expressing the height, h, in terms of the radius, r, using the volume equation for a cylinder:

V = πr^2h

Rearranging the equation to isolate h:

h = V / (πr^2)

Substituting the given volume:

h = (128tπ) / (πr^2)
h = 128t / r^2

Now we can substitute the expression for h in the surface area equation:

SA = 2πr^2 + 2πrh

Substituting the expression for h:

SA = 2πr^2 + 2πr(128t / r^2)

Simplifying the equation:

SA = 2πr^2 + 256t / r

To minimize the surface area, we need to find the critical points where the derivative of the surface area equation is equal to zero.

d(SA) / dr = 0

Differentiating the equation with respect to r:

d(SA) / dr = 4πr - 256t / r^2

Setting the derivative equal to zero:

4πr - 256t / r^2 = 0

Multiplying through by r^2:

4πr^3 - 256t = 0

Simplifying:

r^3 - 64t / π = 0

Solving for r:

r^3 = 64t / π
r = (64t / π)^(1/3)

Now, we can substitute the value of r back into the expression for h:

h = 128t / r^2

Substituting the value of r:

h = 128t / [(64t / π)^(1/3)]^2
h = 128t / [(64t / π)^(2/3)]
h = 128t / [(64^2 / t^(2/3))^(1/3)]
h = 128t / [(64^2)^(1/3) / t^(2/9)]
h = 128t / (4 / t^(2/9))
h = 32t^(11/9)

Therefore, the dimensions of the can with minimum surface area are:

radius, r = (64t / π)^(1/3)
height, h = 32t^(11/9)

To find the dimensions of the can with minimum surface area, we need to optimize the surface area function by finding its critical point(s).

Let's start by expressing the surface area function, SA, in terms of a single variable. We can use the volume formula to eliminate one variable. We know that the volume of the cylinder is 128tπ cm^3, so we can write:

V = πr^2h = 128tπ

Solving this equation for h, we get:

h = (128t) / (πr^2)

Now, substitute this expression for h into the surface area function:

SA = 2πr^2 + 2πrh = 2πr^2 + 2πr[(128t) / (πr^2)] = 2πr^2 + (256t / r)

Next, we need to find the derivative of the surface area function with respect to r:

d(SA) / d(r) = 4πr - (256t / r^2)

To find the critical point(s), we set the derivative equal to zero and solve for r:

4πr - (256t / r^2) = 0
4πr = (256t / r^2)
4πr^3 = 256t
r^3 = (64t) / π
r = (64t / π)^(1/3)

Now that we have the value of r, we can substitute it back into the equation for h to find the corresponding height:

h = (128t) / (πr^2)
h = (128t) / (π[(64t / π)^(1/3)])^2
h = (128t) / ([64t / π]^(2/3))
h = (128t * π^(2/3)) / [64t]^(2/3)
h = (128t * π^(2/3)) / [4^(2/3) * t^(2/3)]
h = 8π^(2/3) / 2^(2/3) t^(1/3)
h = 8π^(2/3) / ∛(4) t^(1/3)

Therefore, the dimensions of the can with minimum surface area are:

Radius, r = (64t / π)^(1/3)
Height, h = 8π^(2/3) / ∛(4) t^(1/3)