The demand function for a product is p=32−8q where p is the price in dollars when q units are demanded. Find the level of production that maximizes the total revenue and determine the revenue. q= units R= $

To find the level of production that maximizes the total revenue, we need to determine the quantity at which the demand function is maximized.

Given that the demand function is p = 32 - 8q, we can rearrange it to solve for q:

p = 32 - 8q
8q = 32 - p
q = (32 - p) / 8

Now, to find the level of production that maximizes revenue, we need to find the quantity that maximizes the product of price (p) and quantity (q), which is the revenue (R).

R = p * q
R = p * [(32 - p) / 8]
R = (32p - p^2) / 8

To determine the revenue, substitute the value of q obtained from the demand function into the revenue equation:

R = p * [(32 - p) / 8]
R = (32p - p^2) / 8

Since we do not know the value of p, we can find the value of p that maximizes the revenue by taking the derivative of the revenue function with respect to p and setting it equal to zero:

dR/dp = [32 - 2p] / 8
0 = 32 - 2p
2p = 32
p = 16

Substitute this value of p back into the revenue equation to find the revenue at this point:

R = (32p - p^2) / 8
R = (32*16 - 16^2) / 8
R = (512 - 256) / 8
R = 256 / 8
R = 32

Therefore, the level of production that maximizes the total revenue is q = (32 - p) / 8 = (32 - 16) / 8 = 2 units, and the revenue at this level is $32.

q=39 r =$41