MHB S8.3.7.3. whose sum is a minimum

[karush]

S8.3.7.3.
Find two positive numbers whose product is 100 and whose sum is a minimum
$x(100-x)=100x-x^2=100$

So far

Looks like it's 10+10=20Doing all my lockdown homework here
since I have no access to WiFi and a PC.
and just a tablet where overkeaf does not work

 

[Prove It]

S8.3.7.3.
Find two positive numbers whose product is 100 and whose sum is a minimum
$x(100-x)=100x-x^2=100$

So far

Looks like it's 10+10=20Doing all my lockdown homework here
since I have no access to WiFi and a PC.
and just a tablet where overkeaf does not work

The way you have defined the product implies that $\displaystyle x + y = 100 $, which it almost certainly doesn't.

You are told the product is 100, so $\displaystyle x\,y = 100 \implies y = \frac{100}{x} $.

The sum needs to be minimised, so your sum function is

$\displaystyle \begin{align*} S &= x + y \\ S &= x + \frac{100}{x} \end{align*} $

Now minimise the amount.

 

[karush]

$S'=\left(x + \dfrac{100}{x}\right)' = 1 - \dfrac{100}{x^2}$
So
$S'(0)=10$

 

[skeeter]

$S'=\left(x + \dfrac{100}{x}\right)' = 1 - \dfrac{100}{x^2}$
So
$S'(0)=10$

No, $S'(0)$ is undefined ...

$S'(x) = 0$ at $x=10$