Question

A deli sells 960 sandwiches per day at a price of $8 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? The deli should charge $ for a sandwich to maximize revenue. (Round to the nearest cent as needed.) (B) A different market survey shows that for every $0.20 reduction in the original $8 price, 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. (Round to the nearest cent as needed.)

Answer 1

Let the amount at the end of the day be "A". A = 960 x 8 = 7680

let us decreases the price by 0.1 dollars "n" times.

The new price will then be: 8 -0.1n

It is said that for every decrease in 0.1 dol, 40 more sandwiches are sold. Thus, the increase in number of sandwiches sold is 40 n

The new amount A = (960+40n)(8-0.1n)

From which we get the equation:

$A(n)=-4n^2+224n+7680 \implies \frac{dA}{dn}=-8n+224\newline\newline \frac{dA}{dn}=0\implies n=28\newline\newline \frac{d^2A}{dn^2}=-8 <0$

(since for maxima, we need first derivative to be qual to zero)

Thus, at n =28, we have a maximum.

n= 28 gives us a total decrease of 2.8 dollars.

b)

Applying the same procedure as above, we get the new amount after the changed price as:

A = (960+5n)(8-0.2n)

$A(n)=-n^2+152n+7680\implies \frac{dA}{dn}=-2n-152 \newline\newline\frac{dA}{dn}=0\implies n=-76 \newline\newline \frac{d^2A}{dn^2}=-2 <0\therefore \text{ maximum}$

So we have n= -76, which means we have to INCREASE the price to get the best sale at the end of the day. The increase is 15.2 dollars