You have a university that costs $c$ to run per semester, making $p$ tuition dollars per student. Whenever the administration raises tuition by $5x\%$, the number of students $n$ goes down by $2x\%$. As things stand, the profit margin ($m = \frac{pn-c}{pn}$) looks bad: it lies at $-20\%$. By how much should the administration change tuition in order to break even (i.e., so that the profit margin equals $0\%)$?
My approach was to write the profit margin in terms of the initial tuition ($p_i$) and current students enrolled ($n_i$) and work out an explicit relationship between tuition, number of students and cost:
$$\frac{p_in_i-c}{p_in_i} = -0.2$$
$$\implies1.2p_{i}n_{i}= c$$
Now I have figured out what the cost is, in terms of the current situation. We want to change the current tuition so that the new profit margin is zero. That is, we would like the following relationship to be true:
$$m = \frac{p_{\text{new}}n_{\text{new}} - c}{p_{\text{new}}n_{\text{new}}} = 0$$
The only time this relationship is true is if: $$p_{new}n_{new} = c = 1.2p_{i}n_{i}$$
If we change the tuition by $x$ percent, the ideal situation is where:
$(p_i(1 + x))(n_i(1 - \frac{2}{5}x)) = c = 1.2 p_in_i$
All we have to do is solve for $x$. The answer is $x = 0.5$ or $x = 1$, under the condition that $n \neq 0$. This is correct according to the solution manual. But I don't know why it's correct.
Lets start again from $1.2p_in_i = c$. Every time $p$ changes, $n$ also changes. In particular, if $p$ increases by $5\%$ (i.e., if $p_{new} = p_{i} + 0.05p_{i}$), then $n$ will drop by $2\%$ (i.e., $n_{new} = n_{i} - 0.02n_{i}$).
We express a multiple of $5\%$ as $0.05x$, and the same multiple of 2% as $0.02x$. So if the price increases by \emph{three} multiples of 5% (i.e., if price increases by 15%), we expect that $n$ will also decrease by \emph{three} multiples of 2% (i.e., quantity sold drops by 6%). Mathematically, we can write:
$p_{new} = p_{i} + 0.05xp_{i} \implies n_{new} = n_{i} - 0.02xn_{i}$
Lets summarize what we have so far. We wanted to find $p_{new}$ so that our profit margin breaks even. This means finding $p_{new}$ so that $p_{new}n_{new} = 1.2 p_in_i$.
$$\implies (p_{i} + 0.05xp_{i})(n_{i} - 0.02xn_{i}) = c$$
And we know that $c = 1.2 p_{i}n_{i}$
$\implies (p_{i} + 0.05xp_{i})(n_{i} - 0.02xn_{i}) = 1.2 p_{i}n_{i}$
Solving the above for $x$ does not yield the correct answer. What did I do wrong?
The difference between the solution from your book and yours is just a matter of scale.
Let's start from your solution:
$$(pi+0.05 y p_i)(ni−0.02 y n_i)=1.2p_in_i $$
I expressed this in terms of $y$ instead of $x$, so we can see how "your" version relates to the book version.
We can first divide both sides by $p_i n_i$, to simplify things:
$$(1 + 0.05 y)(1 - 0.02 y) = 1.2$$ Expanding: $$1 + 0.03 y - 0.001y^2 = 1.2$$ Rearranging, and multiplying by 1,000: $$y^2 - 30 y + 200 = 0$$ This is a quadratic equation with solutions $y = 10$ and $y = 20$.
These correspond to $x= 0.5$ and $x=1$, respectively. In the text book, they use the relative change (not expressed in percent); you used multiples of $5\%$. So indeed, $x = 20 y$, since $1 = 20 \times 5 \%$.
If you want to break even, you need to either double the rate ($x = 1$) or increase it by $50\%$ ($x = 0.5$). Those correspond to $20 \times 5\%$ and $10 \times 5\%$, using your scaled version.