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.


Learn more about this topic:

Loading...
Using Differentiation to Find Maximum and Minimum Values

from Math 104: Calculus

Chapter 9 / Lesson 4
52K

Related to this Question

Explore our homework questions and answers library