Precal Help!!!!!!!!!!!!!!1?

Question

You are desgining a soft drink container that has the shape of a right circular cylinder. The container is to hold 16 fluid ounces. Find the dimensions that will use minimum amount of construction material?

Answer 1

Since you called this precalc, I assume you want to solve without resorting to calculus.

Let
V = volume
S = surface area
r = radius
h = height

We have

V = πr²h = 16 fl oz = 28.875 in³
S = 2πr² + 2πrh

Solve for h in terms of the other variables.

V = πr²h
h = V / (πr²)

Plug into the formula for surface area.

S = 2πr² + 2πrh
S = 2πr² + 2πr[V/(πr²)] = 2πr² + 2V/r

Plug in the value for V.

S = 2πr² + 2V/r = 2πr² + 2*28.875/r = 2πr² + 57.75/r

Without calculus you can only approximate the solution. Use your graphing calculator and see what the approximate minimum value is for r. Then solve for h.
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With calculus I would continue as follows:

S = 2πr² + 2V/r

Take the derivative and set equal to zero to find the critical points.

dS/dr = 4πr - 2V/r² = 0
4πr = 2V/r²
2πr = V/r²
r³ = V/(2π)
r = [V/(2π)]^(1/3)

Take the second derivative to discover the nature of the critical points.

d²S/dr² = 4π + 4V/r³

Plug in the value for r³

4π + 4V/r³ = 4π + 4V/[V/(2π)] = 4π + 8π = 12π > 0
Therefore it is a relative minimum which is what we want.

Solve for h in terms of r.

h = V / (πr²)
h = V / {π [V/(2π)]^(2/3)}
h = [4V/π]^(1/3)
h = [8V/(2π)]^(1/3) = 2[V/(2π)]^(1/3) = 2r

No matter what volume the right cylinder is, the minimum surface area to enclose a given volume will always be achieved when h = 2r.

Now solve for r.

r³ = V/(2π) = 28.875/(2π) = 14.4375/π

r = [14.4375/π]^(1/3) ≈ 1.6625729 in
h = 2r ≈ 3.3251459 in

Answer 2

volume = pi*r^2*H = 16
surface area f(r) = 2pi*r^2+2pi*rH = 2pi*r^2+2(16/r)
f'(r) = 0 =>4pi*r-32/r^2 = 0, r = 2/pi^(1/3), H = 4/pi^(1/3) = 2r
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To get the unit in inches, you can use 16 fluid ounces = 28.875 in^3. Therefore,
r*/r = (28.875/16)^(1/3)
Solve for r*,
r* = 1.66257 in
H* = 2r* = 3.32514 in.

Answer 3

write two equations of volume and area of the cilinder:

V = pi x Square(R) x h
S = 2 x pi x R x h +2 x pi x R^2

Eliminate h from the above two equations, write S as a function of R, take derivative S' and set to zero, solve for R, substitute to have h.

Answer 4

V = pi * h * r^2
16 fl oz * 1.80469 in^3/fl oz = pi * h * r^2
h = 28.875/(pi * r^2)

A = 2 * pi * r^2 + 2 * pi * r * h
A = 2 * pi * r^2 + 2 * pi * r * 28.875/(pi * r^2)
A = 2 * pi * r^2 + 57.75 * r^-1
dA/dr = 4 * pi * r - 57.75 * r^-2 = 0
4* pi * r = 57.75 * r^-2
r^3 = 57.75/(4 * pi)
r^3 = 4.60
r = 1.66 in

h = 3.24 in

Answer 5

16 fluid ounces = 28.875 in^3
V = pir^2h = 28.875 in^3
h = 28.875/pir^2
A = pir^2 +2pirh
A = pir^2 +(2pir)(28.875/pir^2)
A = pir^2 + 57.75/r
dA/dr = 2pir -57.75/r^2
Set dA/dr = 0 getting:
2pir^3 -57.75 = 0
r = (57.75/2pi)^(1/3) = 2.095 inches
h = 28.875/(pi*2.095^2) = 2.095 inches
radius = height