Time remaining:
Please help me solve C and D Please!!!

Mathematics
Tutor: None Selected Time limit: 0 Hours

Mar 30th, 2015

p = (-1/3)x + 100 , 0<= x <= 300 

Revenue = price x quantity sold 
R = px = [(-1 3)x +100]x 
a)   R(x) = -x^2/3 +100x 

b) If 100 units were sold: 
R(x) = -100^2 / 3 + 100(100) = $6,666.67

c)  R(x) = -x^2/3 +100x 

differentiate w.r.t  x
R'(x) = -2x/3 +100 = 0 
             2x/3 = 100 
               2x=300 
                x=150 

A quantity of x=150 maximizes the revenue 
beccause [R''(x) = -2/3 < 0] 


d) R(150) = -x^2/3 +100x  = -22500/3 + 100*150 = -7500+15000 = $7500

Maximum Revenue will be at x = 150   is $7500

Mar 30th, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Mar 30th, 2015
...
Mar 30th, 2015
Mar 27th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer