Question

I have to find the maximum profit in this question, and I keep getting it wrong. I've got the maximum revenue and graph down, but I'm confused. Please see attachment.

Answer 1

@phi
@radar

Answer 2

for the first part you want to find a graph with a cost slope a positive, that eliminates graph labeled B, now the nonlinear function for Revenue will be at 0 when x=0, that leaves only C.

Answer 3

Which I answered correctly. I also answered the second part correctly.
It says Max Profit is revenue minus cost, but who was cost?

Answer 4

P(x) = R(x) - C(x) = -x^2 + 8x - (3x+4) = -x^2 + 8x - 3x -12 = -x^2 + 5x -12
P(x) = -x^2 +5x -12.
P'(x) = -2x + 5 Is this the method you used to find maximum profit?

Answer 5

Not at all. I didn't look at it that way. I was looking at it as Revenue being 16. So 16 minus some cost

Answer 6

How did you get revenue as 16, was it given? I see where someone got 16 as an answer for the b. part. are you going to use that?

Answer 7

No. I got it based off of the graph. Since Revenue is a quadratic equation, maximum revenue is the vertex. The vertex is (4,16), so I got 16 as maximum revenue

Answer 8

O.K. that would be true. That maximum revenue occurs when the widget number 4. What else are we looking for?

Answer 9

Maximum profit. Which would mean we have to find cost. But I get 16 with cost as well, and 16-16=0, so profit would be 0, which is wrong

Answer 10

My goodness, that makes it look like maximum profit is zero!

Answer 11

Which keeps coming out wrong, I'm not sure how to go about this

Answer 12

I've got the correct answer if that helps you derive some sort of meaning
I just don't understand how

Answer 13

When you get the derivative of the profit function P'(x) = -2x + 5, that is a linear equation and when set to 0, you get x = 2 1/2 widgets which is befuddling me I don't think I can help you.

Answer 14

What is the correct answer?

Answer 15

The funny thing is, that's the correct answer

Answer 16

Actually wait, it's 2.25

Answer 17

I don't understand at all

Answer 18

Sorry I don't either. Suggest you repost it and tag someone who you think can.

Answer 19

Alright thanks anyways

Answer 20

@phi
@wio
@whpalmer4

Answer 21

I'm back after reviewing my posts I believe I now have the answer:
Do you understand how I got P(x) = -x^2 + 5x - 4 ??

Answer 22

Yeah I was just doing that, and the vertex states that it would be 2.5, but the answer was 2.25

Answer 23

Yeah I was doing that, and the vertex states that it's 2.5, but the answer is 2.25

Answer 24

From that P'(x) = -2x + 5 set this to 0
-2x + 5 = 0
2x=5
x = 5/2 widgets for maximum profit. Now we substitute 5/2 or (2 1/2) in the P(x) the profit function.
-(5/2)^2 + 5 (5/2) -4
-25/4 + 12 1/2 -4 = -25/4 + 25/2 -4
-25/4 50/4 -16/4 = 9/4 = 2 1/4 or 2.25 I failed to put the value in the profit function and solve. lol

Answer 25

How did you get -2x+5 though?

Answer 26

@Luce did you follow /the -2x + 5 is the derivative of the Profit function which was calculated by R(x) - C(x) and is P(x) -x^2 + 5x -4.

Answer 27

I understand the -x^2+5x-4, but how did you get -2x+5 from that? Do you mean derivative as a normal word, or is that a mathematics term?

Answer 28

Used the power rule for derivative. First term -1x^2 becomes 2(-1)x^(2-1). In words take the exponent and multiply the coefficient. Now reduce the exponent by 1 get -2x. Did you understand, ......have you got into differentiation yet?

Answer 29

Looks like I haven't. Although I'm not sure how I'm supposed to get the answer if I haven't. Would there have been another method?

Answer 30

2nd term 5x is the same as 5x^1, to differentiate (1)(5) x^(1-1) = 5x^0 = 5(1) = 5

Answer 31

Well I guess you could of graphed the -x^2 + 5x - 4 or computed the vertex,

Answer 32

Which is what I did. I just didn't plug it in. Everything makes so much more sense now

Answer 33

Thanks a bunch, Radar

Answer 34

Well you were not the only one who forgot to plug it in the function lol

Answer 35

You're welcome and good luck Luce, I believe you will do well when you do get into calculus.

Answer 36

Thanks, have a nice evening!

Answer 37

You too, I'm outta here.

Answer 38

\[R(x) = -x^2 + 8x\]\[C(x) = 3x+4\]\[P(x) = R(x) - C(x) = -x^2 + 8x - (3x+4) = -x^2 + 5x -4\]
\(P(x)\) is a parabola, in standard form \(y = ax^2 + bx + c\), with \(a = -1,\,b=5\,,c=-4\)
Vertex of a parabola in standard form is found at(*) \[x = -\frac{b}{2a}\]
Therefore the point of maximum profit is at \(x = -\frac{5}{2(-1)} = 2.5\)
At the value \(x = 2.5, \,\,P(x) = -(2.5)^2 + 5(2.5) -4 = -6.25+12.5-4 = 2.25\)

Note that the maximum profit did not occur at maximum revenue!

I've attached a graph of all three quantities for your viewing pleasure.

(*) Note that this is what @radar did with the first derivative of the curve. Hopefully you can follow enough of what he did to see that for a parabola this will always give you the vertex. The calculus approach will allow you find the interesting spots on considerably more complicated graphs, of course, but for a parabola, this is all you need.

Answer 39

Thanks @whpalmer4 , I couldn't remember the x=-b/2a for locating the vertex. An excellent break down of how to approach that problem.

Answer 40

@radar Just have to remember the standard form for a parabola is
\[y = ax^2+bx+c\]First derivative is then \[y' = 2ax + b\]Set that equal to 0 at the vertex\[0 = 2ax + b\]Solve for \(x\)\[-b = 2ax\]\[x=-\frac{b}{2a}\]Now if you ever forget, you can derive it again for the poor souls who have not yet learned the glories of calculus :-)

Answer 41

@whpalmer4 You bet, I know how to develop the quadratic formula using "completing the square" will add this to my bag of tricks. It has been so long since I had calculus that I hesitate to help in that area, especially integrals and differential equations.