# 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?

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

#### Explanation:

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

$x y = 750 \setminus \implies y = \frac{750}{x}$

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

$S = x + 10 y$

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

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

$\setminus \frac{d}{\mathrm{dx}} S = \setminus \frac{d}{\mathrm{dx}} \left(x + \frac{7500}{x}\right)$

$\setminus \frac{\mathrm{dS}}{\mathrm{dx}} = 1 - \frac{7500}{x} ^ 2$

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

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

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

$x = \setminus \pm 50 \setminus \sqrt{3}$

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

${\left(\setminus \frac{{d}^{2} S}{{\mathrm{dx}}^{2}}\right)}_{x = 50 \setminus \sqrt{3}} = \frac{15000}{50 \setminus \sqrt{3}} ^ 3 > 0$

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

$\setminus \implies y = \frac{750}{x}$

$= \frac{750}{50 \setminus \sqrt{3}}$

$= 5 \setminus \sqrt{3}$

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