Answer to Question #192473 in Calculus for Ok099

Question

find two positive numbers whose product is 64 and whose sum is a minimum

Answer 1

Let the numbers be x and y

According to question-

$xy=64\Rightarrow y =\dfrac{64}{x}~~~~~~~-(1)$

Also Let S be their sum-

$S=x+y \\[9pt] S=x+\dfrac{64}{x}$

Differentiate w.r.t. x-

$\dfrac{dS}{dx}=1-\dfrac{64}{x^2}$

Putting $\dfrac{dS}{dx}=0\Rightarrow 1-\dfrac{64}{x^2}=0\Rightarrow x^2=64\Rightarrow x=\pm 8$

Again differentiating $\dfrac{dS}{dx}$ w.r.t. $x-$

$\dfrac{d^2S}{dx^2}=\dfrac{128}{x^3}$

$\dfrac{d^2S}{dx^2}_{\text{ at } x=8}=\dfrac{128}{8^3}=2$

So $\dfrac{d^2S}{dx^2}>0$

SO, S is miniumum at x=8

Then Other number $y=\dfrac{64}{8}=8$

Hence The two numbers are 8 and 8.