How do you find two positive numbers whose product is 750 and for which the sum of one and 10 times the other is a minimum?

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Answer 1 · 2018-07-12T21:33:24

$50 \sqrt{3}$ $50\sqrt3$ & $5 \sqrt{3}$ $5\sqrt3$

Let the positive numbers be $x$ $x$ & $y$ $y$ such that

$x y = 750 \Rightarrow y = \frac{750}{x}$ $xy=750\implies y=750/x$

Let $S$ $S$ be the sum of $x$ $x$ & $10$ $10$ times $y$ $y$ then we have

$S = x + 10 y$ $S=x+10y$

$S = x + 10 (\frac{750}{x})$ $S=x+10(750/x)$

$S = x + \frac{7500}{x}$ $S=x+7500/x$

$\frac{d}{d x} S = \frac{d}{d x} (x + \frac{7500}{x})$ $\frac{d}{dx}S=\frac{d}{dx}(x+7500/x)$

$\frac{d S}{d x} = 1 - \frac{7500}{x^{2}}$ $\frac{dS}{dx}=1-7500/x^2$

$\frac{d^{2} S}{{d x}^{2}} = \frac{15000}{x^{3}}$ $\frac{d^2S}{dx^2}=15000/x^3$

for minimum value of $S$ $S$ we have $\frac{d S}{d x} = 0$ $\frac{dS}{dx}=0$ as follows

$1 - \frac{7500}{x^{2}} = 0$ $1-7500/x^2=0$

$x = \pm 50 \sqrt{3}$ $x=\pm50\sqrt3$

But $x, y > 0$ $x, y>0$ therefore we have $x = 50 \sqrt{3}$ $x=50\sqrt3$ . Now, we have

${(\frac{d^{2} S}{{d x}^{2}})}_{x = 50 \sqrt{3}} = \frac{15000}{{(50 \sqrt{3})}^{3}} > 0$ $(\frac{d^2S}{dx^2})_{x=50\sqrt3}=15000/(50\sqrt3)^3>0$

hence, the sum $S$ $S$ is minimum at $x = 50 \sqrt{3}$ $x=50\sqrt3$

$\Rightarrow y = \frac{750}{x}$ $\implies y=750/x$

$= \frac{750}{50 \sqrt{3}}$ $=750/{50\sqrt3}$

$= 5 \sqrt{3}$ $=5\sqrt3$

Hence, the positive numbers are $50 \sqrt{3}$ $50\sqrt3$ & $5 \sqrt{3}$ $5\sqrt3$