Question

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

Answer 1

Answer:

The required numbers are 8,8

Step-by-step explanation:

Let two positive numbers be x and y

Product of two numbers be xy=64

According to question xy=64

∴y= 64/x

​sum of two positive numbers z=x+y

s=x+ $\frac{64}{x}$

​differentiate with respect to x

$\frac{ds}{dx}$ = $\frac{d}{dx}$(x+$\frac{64}{x}$)

$\frac{ds}{dx}$ = (1-$\frac{64}{x^{2} }$) .... (1)

1-$\frac{64}{x^{2} }$ = 0

$x^{2}$ = 64

x=8

differentiate  the equation (1) again

$\frac{d^{2} s }{dx^{2} }$ = 0 + $\frac{128}{x^{2} }$

∴ at point x=8,

$\frac{d^{2} s }{dx^{2} }$ = $\frac{128}{8^{2} }$

=1/4>0

∴ value of s minimum at point x=8

y = $\frac{64}{x}$

y = 64/8

y = 8

∴ required numbers are 8,8

Hence the explanation.

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Answer 2

The two obtained positive numbers are 8,8.

Given:

The product of the two positive numbers is 64.

The sum is minimum.

Explanation:

Let the two positive numbers be $x$ and $y$.

According to the question,

Product of two numbers be $x y=64$.

determine $y$ in terms of  $x$ .

$\ y=64 / \mathrm{x}$

The sum of two positive numbers, say $s=x+y$

$\mathrm{s}=\mathrm{x}+\frac{64/x}$

Now, differentiate sum with respect to  $x$ .

$\begin{aligned}&\frac{d x}{d x}=\frac{d}{d x}\left(\mathrm{X}+\frac{64}{x}\right) \\&\frac{d x}{d x}=\left(1-\frac{64}{x^{2}}\right) \ldots(1) \\&1-\frac{64}{x^{2}}=0 \\&x^{2}=64 \\&\mathrm{x}=8\end{aligned}$

differentiate the equation (1) again

$\frac{d^{2}s }{dx^{2} } =0+\frac{128}{x^{2} }$

As, $\frac{1}{4}$ greater than $0$.

Therefore the minimum value of sum will occur at  $x$ = $8$.

$y=\frac{64}{x}$

$y=\frac{64}{8}$

$y=8$

Thus, the numbers with minimum sum are 8,8.

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