Optimizing Demand Function: Find Units Sold & Price Per Unit | Help Needed

[JBauer]

The demand function for a product is given by

x = D(p) = 1450 - 3p^3, 0<p<7.85 (please note these should both be less-than-or-equal-to signs)

(a) Find the number of units sold when the price per unit is $5.75.

D(5.75) = 1450 - 3(5.75)^3 = 879.67

Did I set this up correctly? Can anyone confirm this answer?


(b) Find the price per unit when 1000 units are sold.

How would I set this up?

Thanks!

J.B.

 

[stunner5000pt]

i'm really out of touch with this but


your first one appears right. The number of units has to be an integer which you should round down which will yield 879 units, because you cannot sell a fraction of a product!

As for the second one when 1000 units are sold then the D(p) = 1000 and you have to find the value of p then

1000 = 1450 - 3p^3

and solve for p

 

[wais]

(a) Yes, you have set it up correctly. The number of units sold when the price per unit is $5.75 is 879.67.

(b) To find the price per unit when 1000 units are sold, we can use the inverse of the demand function. The inverse demand function is given by p = D^-1(x) = (1450 - x/3)^(1/3). So, we can plug in x = 1000 and solve for p.

p = (1450 - 1000/3)^(1/3) = $5.24

Therefore, the price per unit when 1000 units are sold is $5.24.