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You are in business and have a product that sells for $10 that 1000 people buy each month. You want to make a profit and wonder whether you should raise the price in increments of a dollar. For each dollar you raise the price, you lose 100 customers. How many dollars can you go up to maximize your profit, or should you go up at all?

User MilanHrabos
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1 Answer

18 votes
18 votes

Let:

P = Profit

n = Number of increments by $1

R = Revenue

The profit is given by:


\begin{gathered} P=10+1n \\ P=10+n \end{gathered}

Therefore, the revenue is:


R=(10+n)(1000-100n)

Expand the equation using the distributive property:


\begin{gathered} R=10000-1000n+1000n-100n^2 \\ R=-100n^2+10000 \end{gathered}

The maximum profit of this quadratic function is located at the vertex, we can find the vertex as follows:


\begin{gathered} V(h,k) \\ h=-(b)/(2a) \\ a=-100 \\ b=0 \\ k=R(h)=R(0)=10000 \end{gathered}

The vertex is (0,10000) in another words the maximum profit is achieve if you don't raise the price.

You are in business and have a product that sells for $10 that 1000 people buy each-example-1
User Cabbage Soup
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