2 Answers

Pawel Czuczwara · Answer 1 · 2020-03-19T04:20:45+0000

> Let
be the height of the ad, and
its width. The company wants
h=2w .

> Regardless of the size of the ad, the newspaper will charge $50 minimum. Then for every additional $10 for every square inch. This means the price
of the ad, as a function of the ad's size/area
, is

p(s)=50+10s

Assuming the ad is rectangular, its size/area is given by
s=hw=2w^2 , so we can write the price as a function of the ad's width:

$p(s)=p(2w^2)=50+10(2w^2)\implies P(w)=50+20w^2$

where
P(w) is another price function, but one that depends on
directly (*not* the same as
p(s) , but represents the same thing).

> The company wants the price to be no greater than $2050.

So what we're doing is maximizing the size of the ad,
2w^2 , subject to the price constraint,
$50+20w^2\le2050$ .

- - -

Without using calculus (and I won't bother demonstrating the method that does use it): taking the constraint inequality, we can solve for
to get an idea of what values of
are allowed.

$50+20w^2\le2050\implies20w^2\le2000\implies w^2\le100\implies√(w^2)=|w|\le10\implies-10\le w\le10$