# A deli sells 480 sandwiches per day at a price of $6 each. (A) A market survey shows that for...

## Question:

A deli sells 480 sandwiches per day at a price of $6 each.

(A) A market survey shows that for every $0.10 reduction in the price, 40 more sandwiches will be sold. How much should the deli charge in order to maximize revenue?

(B) A different market survey shows that for every $0.20 reduction in the original $6, 5 more sandwiches will be sold. Now how much should the deli charge in order to maximize revenue?

Now the deli should charge $_____ for a sandwich to maximize revenue.

## Maximizing Revenue:

Problems like the one above are quite common in economics, fortunately, there is a fairly simple straightforward way to solve them. First, we need to *think* a bit about what is actually happening so that we can get a revenue equation, then we can use differentiation to find the max revenue.

## Answer and Explanation:

**Part A**

Right now we sell 480 sandwiches when they are priced at $6 each. Now, we know that for every 10 cent drop in price, we sell an additional 40 sandwiches. This means that if we sell {eq}x {/eq} more sandwiches, then the new Number of Sandwiches Sold {eq}= 480 + 40x {/eq}. We measure price in cents to write the new Price Per Sandwich {eq}= 600 - 10x {/eq}. Then

{eq}\begin{align*} R &= \left( \text{Number of sandwiches sold} \right) \cdot \left( \text{price per sandwich} \right) \\ &= (480 + 40x)(600-10x) \\ &= 40(12+x) \cdot 10 (60 -x) \\ &= 400 (720 + 48x - x^2) \end{align*} {/eq}

This is a downward open parabola and thus has a global max. We differentiate and set equal to 0 to find that it happens when

{eq}\begin{align*} \frac{d}{dx} \left( 400 (720 + 48x - x^2) \right) &= 0 \\ 400 ( 48 - 2x ) &= 0 \\ 2x &= 48 \\ x &= 24 \end{align*} {/eq}

Then the optimal selling price is {eq}600- 10(24) = 360 {/eq} cents, i.e.

{eq}\begin{align*} \$3. 60 \end{align*} {/eq}

**Part B**

The procedure is exactly the same as above. This time we have

{eq}\begin{align*} R &= (480 + 5x)(600-20x) \\ &= 100 (2880 - 66x - x^2) \end{align*} {/eq}

And so the maximizing {eq}x {/eq} is

{eq}\begin{align*} \frac{d}{dx} \left( 100 (2880 - 66x - x^2) \right) &= 0 \\ 100 ( -66 - 2x) &= 0 \\ 2x &= -66 \\ x &= -33 \end{align*} {/eq}

The deli should keep its original price which is $6.

Note that this means that to maximize the revenue, we want to *increase* the price! The maximizing price per sandwich is now

{eq}\begin{align*} 600 - 20(-33) = 1260 \end{align*} {/eq}

which means we want to charge

{eq}\begin{align*} \$12.60 \end{align*} {/eq}

per sandwich. This seems pretty absurd, but when we compare the claim made here to the one in Part a, it itself is absurd: a reduction of ten cents sells an additional 40, but a reduction of twice as much only sells an additional 5. That doesn't seem to make very much sense at all. So, really, neither should this solution.

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from Math 104: Calculus

Chapter 9 / Lesson 4