Question

Suppose that the cost function for a certain product is 𝐶(𝑥) = 400 + 2𝑥 and the price function is 𝑝(𝑥) = 20 − 0.02𝑥. Find the production level that maximizes profit.

Answer 1

First you need to find profit. Subtract the cost from the price to get profit

Answer 2

So essentially,

\[Profit = Price - Cost \]

\[f(x) = p(x)-c(x)\]

Answer 3

yes

Answer 4

This is of course, assuming cost means cost to the producer, and price means how much money it takes for someone to buy it

Answer 5

20-0.02x - (400+2x)

20-0.02x -400 - 2x

Combining like terms

20-(-400)

0.02x-2x

I get this equation for profit.

y(x) = -2.02x+420

Answer 6

it would be -380, not +420

Answer 7

I know what to do after, you find the derivative and set = to zero

Answer 8

y(x) = -380-2.02x

Answer 9

yes

Answer 10

\[\frac{ dy }{ dx }(-380-2.02x) = -2.02 \]

Answer 11

To me this answer doesn't make much sense.

Answer 12

@hartnn

Answer 13

this doesn't have a largest value
is there supposed to be an \(x^2\) somewhere in the question?

Answer 14

:( well... it didn't say this in the question but I found in my notes that revenue

R(x) = x*p(x)

or revenue = x times the price function

Answer 15

yes .... profit = revenue - cost = x*price - cost

Answer 16

that makes more sense

Answer 17

Okay this is what I got now

\[x*p(x)-C(x) = x*(20-0.02x)-400+2x \]

\[20x-0.02x^{2}-400+2x\]

Answer 18

\[x( 20 − 0.02𝑥)-( 400 + 2𝑥)\] looks better

Answer 19

no, don't forget to distribute the minus sign!

Answer 20

oh

Answer 21

Okay so would it be this?

\[20x-0.02x^{2}-400-2x = -0.02x^{2}+18x-400\]

Answer 22

wondering if I grouped the terms properly

Answer 23

yes

Answer 24

max is at \(-\frac{b}{2a}=-\frac{18}{2\times (-.02)}\)

Answer 25

okay, is that a general formula for any parabolas -b/2a for the maximum?

Answer 26

the way I was taught was you take the derivative and set it = 0

Answer 27

max or min, depending if it opens up or opens down
it is the first coordinate of the vertex

Answer 28

so if it's value is >0 it opens up, maximum <0 down minimum?

Answer 29

on the other hand, the derivative of \[y=ax^2+bx+c\] is \[y'=2ax+b\] set it equal to zero and solve, you always get \[x=-\frac{b}{2a}\]

Answer 30

and no, you have that backwards, if it opens down you have a max, if it open up, you have a min

Answer 31

interesting

-0.04x+18 = 0

-18/-0.04

x = 18/0.04

x = 450 so this is what I got

so I need to put this value back into original equation to find out what the maximum profit.

Answer 32

Oh I see now, the figure you drew clarified it.

Answer 33

good

Answer 34

So now that i got the derivative I take what I get and plug it into the original function we started out with, and this gives us the profit.

-0.02(450)^{2}+18(450)-400 = $3,650

So I also thought to plug in what I got for the other equations too

p(x) = 20-0.02(450) = 11 so the price of each unit is 11 dollars

and the cost

C(x) = 400+2(450) = $1,300


my prof wrote down to test if its a local maximum or minimum

but what's the point of that anyway cant we just find the derivate at zero and plug it back into f(x)?

Answer 35

since you know it is a parabola that opens down, you know it has a max and not a min
not sure what else there is to say

Answer 36

ok

Answer 37

so the profit is 3,650 the unit price is 11 dollars and it will cost 1,300