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I'm reading a book (Calculus Made Easy - Silvanus Thompson) that has the following exercise:

Divide a number N into two parts in such a way that three times the square of one part plus twice the square of the other part shall be a minimum.

It's part of a chapter that explains how to find minima/maxima using second-order derivatives. I'm quite confused on how to solve this problem.

I tried calculating the derivative of the following:

$$y = 3(xN)^2 + 2[(1-x)N]^2$$

Where $x$ represents a certain percentage of $N$. This gave me:

$$\frac{dy}{dx} = 2N(5x-2)$$

Which I then equated to $0$, which gave me the wrong value for $x$ when comparing to the answer key (which is $0.4N, 0.6N$). Clearly I'm doing it wrong.

Any help would be greatly appreciated. Please bear in mind that I'm trying to learn this stuff by myself without any proper math background.

GFlow
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    Shouldn't it be $3(xN)^2$ instead? – player3236 Sep 23 '20 at 14:53
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    $n$ and $N$ are different. You shouldn't switch from one to the other. The problem calls for squaring the first, not cubing, before multiplying by $3$ – Ross Millikan Sep 23 '20 at 14:59
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    Note that if both parts are squared (instead of cubing one part and squaring the other), then a common factor of $N^2$ can be removed (distributed out) of the expression to be minimized. – hardmath Sep 23 '20 at 15:00
  • @player3236 You're right! I changed it. – GFlow Sep 23 '20 at 16:23
  • @hardmath Yes, now it all adds up, I get x = 4/10, which is the answer I'm looking for. Sometimes I get tunnel vision, my bad. – GFlow Sep 23 '20 at 16:25

2 Answers2

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$3x^2+2{(N-x)}^2=5x^2-4Nx+2N^2=5(x-\frac{2N}{5})^2+\frac{6N^2}{5}$

It will get min when $x=\frac{2N}{5}=0.4N$ and the other part is $0.6N$

Lion Heart
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After your edit, it looks correct.

$5x-2=0$ when $x=\frac25=0.4$, so $xN = 0.4 N$ and $(1-x)N=0.6N$.

J. W. Tanner
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