max profit?

[SilentSymphony]

it costs C dollars to manufacture and deliver a product. If the product sells at x dollars each, the number sold is given by

n= [a/(x+c)] + b(100-x)

where a and b are constants. What selling price will bring you a maximum profit?

 

[Deleted member 4993]

Please show us your work - what you have tried even if you think it is wrong and exactly where you are stuck.

 

[SilentSymphony]

okay, i know that im looking for x, and that the easiest way to do that to where i can find the values is to differentiate after i get x by itself.

n = (a/(x-c)) + 100b - bx
bx = (a/(x-c)) +100b - n
bx^2 -bxc = a +100bx-100bc -nx+nc
bx^2 -bxc -100bx+nx = a -100bc +nc
x (bx -bc -100b+n) = a-100bc+nc
x= (a-100bc+nc)/(bx-bc-100b+n)

and that's where i know ive messed up somewhere. im sure that im just making some careless error somewhere, but ive been through it enough to know that i need help. thanks yall!

 

[galactus]

You're making it too tough. Differentiate, then set to 0 and solve for x.

\(\displaystyle \L\\n=\frac{a}{x+c}-bx+100b\)

Use the quotient rule on the fractional part:

\(\displaystyle \L\\\frac{(x+c)(0)-a(1)}{(x+c)^{2}}=\frac{-a}{(x+c)^{2}}\)

You have:

\(\displaystyle \L\\n'=\frac{-a}{(x+c)^{2}}-b\)

Now, set to 0 and solve for x. It ain't bad.