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Math Help - confuse in differentiation application problem

  1. #1
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    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.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by afeasfaerw23231233 View Post
    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
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