> Let [tex]h[/tex] be the height of the ad, and [tex]w[/tex] its width. The company wants [tex]h=2w[/tex].
> 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 [tex]p[/tex] of the ad, as a function of the ad's size/area [tex]s[/tex], is
[tex]p(s)=50+10s[/tex]
Assuming the ad is rectangular, its size/area is given by [tex]s=hw=2w^2[/tex], so we can write the price as a function of the ad's width:
[tex]p(s)=p(2w^2)=50+10(2w^2)\implies P(w)=50+20w^2[/tex]
where [tex]P(w)[/tex] is another price function, but one that depends on [tex]w[/tex] directly (*not* the same as [tex]p(s)[/tex], 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, [tex]2w^2[/tex], subject to the price constraint, [tex]50+20w^2\le2050[/tex].
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Without using calculus (and I won't bother demonstrating the method that does use it): taking the constraint inequality, we can solve for [tex]w[/tex] to get an idea of what values of [tex]w[/tex] are allowed.
[tex]50+20w^2\le2050\implies20w^2\le2000\implies w^2\le100\implies\sqrt{w^2}=|w|\le10\implies-10\le w\le10[/tex]
The width of the ad can't be negative, so the actual interval of allowed values for [tex]w[/tex] would be [tex]0<w\le10[/tex].
Since [tex]2w^2\ge0[/tex] for all [tex]w[/tex], it stands to reason that its maximum value will occur at the end of this interval when [tex]w=10[/tex] inches.
The ad's height is twice its width, so the height of the add would be [tex]h=20[/tex] inches, which makes D the correct answer.
