A closed cylindrical can is to hold 2 litter (200 cm³) of liquid. How should we choose the height and radius to minimize the amount of material needed to manufacture the can?
A litter is a group of kittens, a litre is a fluid measure of 1000 ml
let the height be h cm
let the radius be r cm
πr^2 h = 200
h = 200/(πr^2)
amount of material ---> surface area (SA)
SA = 2πr^2 + 2πrh
= 2πr^2 + 2πr(200/(πr^2))
= 2πr^2 + 400/r
differentiate, set that equal to zero, and solve for r
Plug back into h= ...
and you are done
Radius=1.99735...
Height=3.984...
I am sorry, I put wrong answer
To minimize the amount of material needed to manufacture the can, we need to optimize the dimensions of the cylindrical can. The can should have a minimum surface area while still holding 2 liters (200 cm³) of liquid.
Let's start by finding the volume formula for a cylinder:
V = πr^2h
We need to hold 200 cm³ of liquid, so:
V = 200
Now we need to minimize the surface area, which is given by the formula:
A = 2πr^2 + 2πrh
We can rewrite the volume equation in terms of height and radius by solving for h:
h = V / (πr^2)
h = 200 / (πr^2)
Now let's substitute this value of h into the surface area equation:
A = 2πr^2 + 2πr (200 / (πr^2))
A = 2πr^2 + 400 / r
To minimize the surface area, we can find the derivative of A with respect to r and set it equal to zero:
dA/dr = 4πr - 400 / r^2 = 0
Solving this equation will give us the value of r that minimizes the surface area.