Maximizing Profit

[summer_storms]

The total profit (in dollars) for sales of "x" rowing machines is given by
P(x) = 0.2x^2 + 300x - 200
What is the profit if 500 are sold? For what value of "x" will the profit be at a maximum?


What I have done so far:

P(x) = 0.2x^2 + 300x - 200
Let x = 500
P(500) = 0.2(500)^2 + 300(500) - 200
P(500) = 10,000 + 150,000 - 200
P(500) = 160,000 - 200
P(500) = 159,800
P(500)/500 = 159,800/500
P = 319.60

The profit is $319.60

How would I go about figuring the second part? When the value of x profit will be at a maximum?

Thank you for your time.

 

[tkhunny]

Several Ways. Which do you know?

1) Completing the Square is a beautiful thing.

2) Find the zeros and calculate their average.

3) x = -b/2a

 

[Denis]

P(x) = 0.2x^2 + 300x - 200
Let x = 500
P(500) = 0.2(500)^2 + 300(500) - 200
P(500) = 10,000 + 150,000 - 200 : INCORRECT .2(500)^2 not 10000

 

[summer_storms]

P(x) = 0.2x^2 + 300x - 200
Let x = 500
P(500) = 0.2(500)^2 + 300(500) - 200
P(500) = 10,000 + 150,000 - 200 : INCORRECT .2(500)^2 not 10000


Is it then 50,000?

changing the rest of the equation to

P(500) = 50,000 + 150,000 - 200
P(500) = 200,000 - 200
P(500) = 199,800
P(500)/500 = 199,800/500
P = 399.6 or a profit of $399.60

 

[summer_storms]

Several Ways. Which do you know?

1) Completing the Square is a beautiful thing.

2) Find the zeros and calculate their average.

3) x = -b/2a

I think I know them all at times, other times I'm not so sure of myself, lol.
I'm going to try #3 and see how that works for me:

-300/0.2(2)
-300/0.4
-750
So the profit would be maximized when 750 rowing machines are sold?

 

[soroban]

Hello, summer_storms!

There must be a typo in the problem . . .

The total profit (in dollars) for sales of \(\displaystyle x\) rowing machines is given by:
. . \(\displaystyle P(x)\:=\:0.2x^2\,+\,300x\,-\,200\)

For what value of \(\displaystyle x\) will the profit be at a maximum?

The profit function is an up-opening parabola . . . It has no maximum.

. . - . . . - . . . . - . . . - . . . . . . . . . . . .↓
Assuming that the function is: \(\displaystyle \(x)\:=\:\)-\(\displaystyle 0.2x^2\,+\,300x\,-\,200\)
. . we have a down-opening parabola; its maximum is at its vertex.

The vertex is at: \(\displaystyle \,x\,=\,\frac{-b}{2a}\;\;\) We have: \(\displaystyle \,a\,=\,-0.2,\:b\,=\,300,\:c\,=\,-200\)

The vertex is: \(\displaystyle \,x\:=\:\frac{-300}{2(-0.2)} \:=\:750\)


Therefore, maximum profit is achieved with the sale of 750 rowing machines.

 

[summer_storms]

Hello, summer_storms!

There must be a typo in the problem . . .

The total profit (in dollars) for sales of \(\displaystyle x\) rowing machines is given by:
. . \(\displaystyle P(x)\:=\:0.2x^2\,+\,300x\,-\,200\)

For what value of \(\displaystyle x\) will the profit be at a maximum?

The profit function is an up-opening parabola . . . It has no maximum.

. . - . . . - . . . . - . . . - . . . . . . . . . . . .↓
Assuming that the function is: \(\displaystyle \(x)\:=\:\)-\(\displaystyle 0.2x^2\,+\,300x\,-\,200\)
. . we have a down-opening parabola; its maximum is at its vertex.

The vertex is at: \(\displaystyle \,x\,=\,\frac{-b}{2a}\;\;\) We have: \(\displaystyle \,a\,=\,-0.2,\:b\,=\,300,\:c\,=\,-200\)

The vertex is: \(\displaystyle \,x\:=\:\frac{-300}{2(-0.2)} \:=\:750\)


Therefore, maximum profit is achieved with the sale of 750 rowing machines.

Hi, Soroban. Great catch. Yes I missed the negative sign when inputting my information. That makes a huge difference in the rest of the problem also. Thanks a bunch!