Question

Determine two positive numbers whose product is 360, and the sum of one of the numbers and 10 times the other number is a minimum.

Answer 1

Answer:

$60, 6$

Step-by-step explanation:

Let $x$ be one number and $y$ be the other number.

The product of the two numbers is 360.

$xy=360\\\Rightarrow y=\dfrac{360}{x}$

Sum of one of the numbers and 10 times the other number be z

$z=x+10y\\\Rightarrow z=x+10\times\dfrac{360}{x}\\\Rightarrow z=x+\dfrac{3600}{x}\\\Rightarrow z=\dfrac{x^2+3600}{x}$

Differentiating with respect to $x$ we get

$z'=\dfrac{2xx-(x^2+3600)1}{x^2}\\\Rightarrow z'=\dfrac{2x^2-x^2-3600}{x^2}\\\Rightarrow z'=\dfrac{x^2-3600}{x^2}$

Equating with zero

$0=\dfrac{x^2-3600}{x^2}\\\Rightarrow x=\sqrt{3600}\\\Rightarrow x=60$

Double derivative of z

$z''=\dfrac{2xx^2-2x(x^2-3600)}{x^4}\\\Rightarrow z''=\dfrac{2x^3-2x^3-7200x}{x^4}\\\Rightarrow z''=\dfrac{7200}{x^3}$

at $x=60$

$z''=\dfrac{7200}{60^3}\\\Rightarrow z''=0.033>0$

So $z$ is minimum at $x=60$

$y=\dfrac{360}{x}\\\Rightarrow y=\dfrac{360}{60}\\\Rightarrow y=6$

So, one number is 60 and the other number is 6.