Question

Find two positive numbers satisfying the given requirements. The product is 216 and the sum is a minimum. (No Response) (smaller value) (No Response) (larger value)

Answer 1

Answer:

$x1=\sqrt{216} \ and \ y1=\ \sqrt{216}$

Step-by-step explanation:

Let the first number is x1 and other number is y1 then

$x1 * y1 =216$

Therefore

$y1=$$\frac{216}{x1}$

also there sum is

$s1 =x1+y1$....Eq(1)

Putting the value of y1  in the previous equation

$s1\ =x1 + \frac{216}{x1}$........Eq(2)

Differentiate the the Eq(2) with respect to x1 we get

$\frac{ds1}{dx1} \ =\ 1+216*\frac{1}{-x1^{2} }$

$\frac{ds1}{dx1} \ =\ 1-\frac{216}{x1^{2} }\ =\ 0$

${x1^{2} }\ =\ 216\\ x1=\sqrt{216}$

Putting the value of X1 in Eq(1) we get

$y1=\frac{216}{\sqrt{216} } \\y1=\frac{216*\sqrt{216} }{216} \\y1=\ \sqrt{216}$

So $x1=\sqrt{216} \ and \ y1=\ \sqrt{216}$