Maximum Revenue. A university student center sells 1,600 cups of coffee per day

Question

Maximum Revenue. A university student center sells 1,600 cups of coffee per day at a price of dollar2.40. A market survey shows that for every $0.05 reduction in price, 50 more cups of coffee will be sold. How much should the student center charge for a cup of coffee in order to maximize revenue A different market survey shows that for every dollar0.10 reduction in the original dollar2.40 price, 60 more cups of coffee will be sold. Now how much should the student center charge for a cup of coffee in order to maximize revenue?

Explanation / Answer

Here, we have

y=(2.40 - .05x) (1600 + 50x) = -2.5x2 + 120x -80x + 3840

On differentiating,we get
-5x + 40 = dy/dx

For critical points, put dy/dx=0 to get
-5x =-40
x=8

Now, we plug x=8 in the function to find the max revenue

i.e. -2.5(8)^2 + 40(8) + 3840 = $ 4000


B.

Similarly,

y=(2.40 - .10x) (1600 + 60x) =-6x2 + 16x +3840

On differentiating,we get
dy/dx= -12x - 16

For critical points, put dy/dx=0 to get
-12x - 16 = 0
-12x = 16
x= -4/3

The negative sign shows that profit will never be maximized beyond the original sale.

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---