# Minimizing Surface Area of a Can

1. Jun 15, 2007

### JJV19

This is what i have so far

SA = 2(pi)r^2 + 2(pi)r(245.45 / (pi)r^2)

Derivative of SA = - 490.9r^2 + 4(pi)r

0 = -490.9r^2 + 4(pi)r

and i'm stuck at this step.. i've tried a bunch of things to try and solve for r but i can't seam to get a logical answer for the radius of the minimized pop can

2. Jun 15, 2007

### itsjustme

minimise, thats something new, well to minimise why just not have a can at all? make r=0. this is one of the roots of the equation you gave me and it it the only min turnning point as d^2/dx^2 at the point is greater than 0. Moreso the other root 0.025 has d^2y/dx^2 less than 0 making it a maximum turnning point. The working follows:

0 = -490.9r^2 + 4(pi)r
0=r(-490.9r+4(pi))
hence r=0 or 490.9r=4(pi)
r=0 or r=0.02559

3. Jun 15, 2007

### JohnDuck

You didn't differentiate S correctly. According to your post,

$$S=2\pi r^{2} + 2\pi r(\frac{245.45}{\pi r^{2}}) = 2\pi r^{2} + \frac{490.9}{r}$$

and

$$\frac{dS}{dr}=4\pi r - 490.9r^{2}$$

Do you see where your error is?

4. Jun 15, 2007

### itsjustme

-490.9r^-2 not ^2

5. Jun 15, 2007

### JJV19

thanks alot both of you lol .. it's funny how the simple mistakes often cause the most grief