Question

The product of two numbers is 32. Find the sum of the two numbers S(x) as a function of one of the numbers, x. S(x) = Find the positive numbers that minimize the sum and list them in increasing order: (smaller number) = ?(larger number) = ?Let c be the smaller of the two numbers that minimize the sum. Then c = ?S'(c) = ? and S"(c) = ?

Answer 1

Answer:

$c = 4\sqrt2\\\\S'(4\sqrt2) = 0\\\\S''(4\sqrt2)=\frac{1}{2\sqrt2}$

Step-by-step explanation:

We are given the following information in the question:

The product of two numbers is 32.

Let the two numbers be x and y, then,

$x\times y = 32\\\\y = \displaystyle\frac{32}{x}$

The sum of the two numbers = S(x)

$S(x) = x + \displaystyle\frac{32}{x}$

First, we differentiate S(x) with respect to x, to get,

$\displaystyle\frac{d(S(x))}{dx} = \frac{d( x +\frac{32}{x})}{dx} = 1 - \frac{32}{x^2}$

Equating the first derivative to zero, we get,

$\frac{d(S(x))}{dx} = 0\\\\1 - \frac{32}{x^2}= 0$

Solving, we get,

$x =\pm \sqrt{32}$

Again differentiation S(x), with respect to x, we get,

$\displaystyle\frac{d^2(S(x))}{dx^2} =\frac{64}{x^3}$

At$x =\sqrt{32}$,

$\frac{d^2(S(x))}{dx^2} > 0$

Thus, minima occurs at x = $\sqrt{32}$ for S(x).

If c be the smaller of the two numbers that minimize the sum, then,

$c = \sqrt{32} = 4\sqrt2\\\\S'(c) = 1 - \displaystyle\frac{32}{32} = 0\\\\S''(c) = \frac{64}{32\sqrt{32}} = \frac{1}{2\sqrt2}$