My math skills are getting rusty. I am trying to work out what the formula should be for calculating price, $P$, based on a formula I used to calculate margin, $\mu$, with a parameter, cost, $C$.
$$\mu = 39.32\text{, }\ P = 177.99 \text{, }\ C = 108.00.$$
$$\mu = \frac{100(P-C)}{P}$$
I know $P$ should be $177.99$ when $C$ is $108.00$ and $\mu$ is $32.39$. But what is the formula to calculate it? $$P = F(\mu,C) $$
After going through several math tutorials I was able to simplify the formula a little. But I am still at a loss as to how to get price to one side of the equation by itself.
$$P=\frac{P\cdot\mu}{100} +C$$
3 Answers
Hint:
A nice formatting helps me understanding, perhaps for you too: $$ \mbox{price} = \frac{\mbox{margin}}{100} \mbox{price} + \mbox{cost} \iff \\ \mbox{price} - \frac{\mbox{margin}}{100} \mbox{price} = \mbox{cost} \iff \\ 1 \cdot \mbox{price} + \left(- \frac{\mbox{margin}}{100}\right) \mbox{price} = \mbox{cost} $$ Now use the distributive law $$ A \cdot C + B \cdot C = (A + B) \cdot C $$ for the left hand side, to consolidate terms with price. You just have to match $A$, $B$ and $C$ with the term $$ 1 \cdot \mbox{price} + \left(- \frac{\mbox{margin}}{100}\right) \mbox{price} $$ and apply the law. Start with $C$ because it shows up twice and is the reason why we do this, because we want to have $C$ show up only once, like in the right hand side of the distributive law.
An example: $5 \cdot 7 + 3 \cdot 7 = (5+3)\cdot 7$. thus $A =5$, $B=3$, $C=7$.
Solution:
$$ 1 \cdot \mbox{price} + \left(- \frac{\mbox{margin}}{100}\right) \mbox{price} = \mbox{cost} \iff \\ \left( 1 + \left(- \frac{\mbox{margin}}{100}\right)\right) \mbox{price} = \mbox{cost} \iff \\ \mbox{price} = \frac{\mbox{cost}}{\left( 1 + \left(- \frac{\mbox{margin}}{100}\right)\right)} \iff \\ \mbox{price} = \frac{\mbox{cost}}{\frac{100 -\mbox{margin}}{100}} = \frac{100}{100 -\mbox{margin}} \mbox{cost}$$
Note: This answer uses the variable names from the OP's version of the question.
Hint: Notice that
$$ \frac{\mathrm{price} - \mathrm{cost}}{\mathrm{price}} = \frac{\mathrm{price}}{\mathrm{price}} - \frac{ \mathrm{cost}}{\mathrm{price}} = 1 - \frac{ \mathrm{cost}}{\mathrm{price}} $$
There should only be 1 "price" left after this.
Note that $$\mu = \frac{100(P-C)}P = \frac{100 P - 100 C}P = 100 \frac PP - 100 \frac CP = 100 - 100 \frac CP$$ So rearranging gives $$100\frac CP = 100 - \mu$$ Divide by $100 C$ to get $$\frac1P = \frac{100 - \mu}{100 C}$$ And finally take reciprocals on both sides: $$P = \frac{100 C}{100 - \mu}$$
