Thread: confuse in differentiation application problem

1. confuse in differentiation application problem

i do this question correctly but i am confused about something in (c)
$\frac {dV}{dr}= \frac 5 {12} (S- \frac {20\pi ^2 r^4}S)$
$r = ^4\sqrt {\frac {S^2}{20\pi^2}}$
i don't understand why it shouldn't be
$\frac {dV}{dr}= \frac 5 {12} (S+ r\frac{dS}{dr} -\frac{20\pi^2 r^4}S + \frac {4\pi^2 r^5}{S^2} )$ ?z
why the surface area S is treated as a constant? S is a function of r and h, S changes as r or h changes. Although V is a function of r and S after expressing h in terms of r and S, then S is not a variable?

for example, the perimeter P of a rectangle is given by $P = 2(x+y)$ and its area $A = xy$ , then $y = \frac P 2 -x$ and $A = \frac P 2 x - x^2$ if i differentiate both sides then should it be $\frac {dA}{dx }=\frac P 2 - 2x$ or $\frac {dA}{dx }=\frac P 2 +\frac x 2 \frac {dP}{dx}- 2x$ ?

any explanation will be appreciated! thanks.

2. Originally Posted by afeasfaerw23231233
iwhy the surface area S is treated as a constant? S is a function of r and h, S changes as r or h changes. Although V is a function of r and S after expressing h in terms of r and S, then S is not a variable?
Well, h is related to r and so is not really an independent variable, and you are told that the surface area is S (and by implication is a constant, it could have been worded better). Then you are to find the value of r that maximises V subject to the surface area being S. Without this constraint (that S is fixed) there is no maximum for V, as we can let r become arbitarily large and V will also become arbitarily large.

RonL