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## Homework Statement

The total surface area of a right circular cylinder is given by the formula A = 2pir(r + h) where r is the radius and h is the height.

a) Find the rate of change of A with respect to h if r remains constant.

b) Find the rate of change of A with respect to r if h remains constant.

c) Find the rate of change of h with respect to r if A remains constant.

## Homework Equations

A = 2pir(r + h)

## The Attempt at a Solution

I thought i had this all right but then realized that what i had done for part a was wrong, i think that part b and c are right but i will just put my answers up to check.

a) A = 2pir(r + h)

r = c (constant)

A = 2pic(c + h)

A = 2pi(c

^{2}+ ch)

A/2pi = c

^{2}+ ch

however, i do not know at this point, how i would isolate c!

What i did was bring the A/2pi over and get c

^{2}+ ch - A/2pi = 0 and use the quadratic formula. From that i got that c = (h +- h)/4pi and reason that c cannot be zero or else the surface area would be zero so c must be h/2pi. I think that this process of my work is wrong because i dont think you can use the quadratic formula and solve for c because c is a constant.

We know that

A = 2pic(c + h)

A = 2pic

^{2}+ 2pich

dA/dh = 2pic = 2pi(h/2pi) = h

As you can see i dont think that what i did was correct, please help me with this one, a) is what i need most help with

b) h = c (constant)

A = 2pir(r + c)

A/2pir = r + c

c = (A/2pir) - r

dA/dr = 2pi(r + c) + 2pir = 4pir + 2pic = 4pir + 2pi((A/2pir) - r) = 4pir + A/r - r

c) A = c (constant)

c = 2pir(r + h)

c/2pir - r = h

dh/dr = (-c/2pir

^{2}) - 1

= (-(2pir(r + h))/2pir

^{2}) - 1

= ((- r - h) / r) - 1

= (-2r - h) / r