Calculus problem solving?

Question

A cylindrical storage tank with a capacity of 1000m^3 is to be constructed. The specifications call for the top and base to be made of sheet steel, which costs $100/m^2, and the wall to be made of sheet steel, which costs $80/m^2. Determine the dimensions of the tank that would minimize the cost of construction.

Answer 1

Create an equation for the cost of a cylandar: c=100(3.14r^2)+100(3.14r^2)+80(6.28r*h)

Create a volume equation: 1000=3.14r^2*h
Solve the volume equation for h: h=1000/(3.14r^2) and substitute this for h in the cost equation: c=100(3.14r^2)+100(3.14r^2)+80(6.28r*1000/(3.14r^2))

Simplify the cost equation as it is getting pretty ugly: c=628r^2+160000r^(-1)

Now to find maximums and minimums you need the derivative of the equation, whan a derivative is zero, the original equation has a maximum or minimum.

The derivative=1256r-160000r^-2
Solve for 0: 0=1256r-160000r^-2
1256r=160000r^-2
r^3=127.388
r=5.031

When the radius is 5.031, the cost is minimized. plug this back into the volume to find the height h=1000/(3.14(5.031)^2) which is 12.582

Answer: Costs are minimized when the tanks dimensionas are as follows: Height: 12.582m Radius: 5.031m

Answer 2

From the volume of a cylinder V = r^2*pi*h = 1000 you find h = 1000/(pi*r^2). The cost = 100*2*r^2*pi + 80*2*r*pi*h = 200*r^2*pi + 160000/r. To find minimum cost depending on r you set first derivative on r to 0 i.e.: 400*r*pi = 160000/r^2. Therefore r = 5.031 and h = 12.577.

Answer 3

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

V = πr²h
S = 2πr² + 2πrh

Given
V = 1000 m³
C = (100)2πr² + (80)2πrh = 200πr² + 160πrh

We have

V = πr²h
h = V/(πr²)

Substituting

C = 200πr² + 160πrh = 200πr² + 160πr{V/(πr²)}
C = 200πr² + 160V/r

Take the first derivative and set it equal to zero to find the critical value(s).

dC/dr = 400πr - 160V/r² = 0
400πr = 160V/r²
5πr = 2V/r²
r³ = 2V/(5π)
r = {2V/(5π)}^(1/3)

Take the second derivative to find the nature of the critical value(s).

d²C/dr² = 400π + 320V/r³ > 0 since both terms are positve
This implies a relative minimum, which is what we want.

Solve for h in terms of r

r³ = 2V/(5π)

h = V/(πr²)
h = (2*5)V/{(2*5)πr²} = {2V/(5π)}{5/(2r²)}
h = r³{5/(2r²)} = 5r/2

Solve for r and h.

r = {2V/(5π)}^(1/3) = {2*1000/(5π)}^(1/3) = (400/π)^(1/3)
r = 2(50/π)^(1/3) ≈ 5.030796 m

h = 5r/2 = (5/2){2(50/π)^(1/3)}
h = 5(50/π)^(1/3) ≈ 12.57699 m

Answer 4

this is alot of calculation
2 equations

1000 = r^2 pie h (volume of cylin. circle * height)
..C= (100)(r^2 pie)(2) + 2r pie h (80)
cost....^.........^........^..........^.....^....^
.......$100/m^2 * circle......circumference*height
..........* (top + base)............*$80/m^2

the first equation can be rewrite as
h = 1000 / (r^2 pie)

then substitution:
C= (100)(r^2 pie)(2) + 2r pie [1000 / (r^2 pie)] (80)

simplify
= (200)(r^2 pie) + 160000/ r

derive
C' = 400 r pie - 160000 r^(-2)

set C' = 0
0 = 400 r pie - 160000 r^(-2)
160000 r^(-2) = 400 r pie
400 r^(-2) = r pie
400 = r^3 pie
400 / pie = r^3
(400 / pie) = r, which is about 5.03m
plug it in for h = 12.58m.

so around h = 12.58m and r = 5.03m.

Answer 5

sin(a million/x)-x*cos(a million/x)(a million/x^2) i imagine it truly is genuine using the product rule then the chain rule on the only ideal section. "spinoff of x cases sin(a million/x) plus x cases the spinoff of sin(a million/x)"