Revenue and Cost and Profit Problem

[needOfHelpCMath]

Given the following revenue and cost functions, find the maximum profit.

R(x) = 71x - 2x^2; C(x) = 23x + 108

P(x) = Revenue - Cost

P(x) = 71x-2x^2 - (23x+108)

P(x) = -2x^2 + 48x - 108 *I am stuck right here. Using quadratic formula but cannot seem to solve it what I am doing
wrong or have I miss any step?

 

[MarkFL]

Your profit function is correct:

\(\displaystyle P(x)=-2x^2+48x-108\)

Now, what I would do is find the axis of symmetry (since the vertex lies on this axis)...recall that for the general quadratic:

\(\displaystyle y=ax^2+bx+c\)

The axis of symmetry is the line:

\(\displaystyle x=-\frac{b}{2a}\)

This is the arithmetic mean of the two roots. What is the axis of symmetry for your profit function?

 

[MarkFL]

Another approach would be to express your profit function in vertex form:

\(\displaystyle P(x)=-a(x-h)^2+k\)

where:

\(\displaystyle P_{\max}=k\)

 

[MarkFL]

To follow up:

The axis of symmetry is:

\(\displaystyle x=-\frac{48}{2(-2)}=12\)

We see that the profit function opens downward, and so the vertex is the global maximum, given by:

\(\displaystyle P_{\max}=P(12)=-2(12)^2+48(12)-108=12(-24+48-9)=12\cdot15=180\)

If we use the vertex form approach, we may write:

\(\displaystyle P(x)=-2(x^2-24x)-108=-2(x^2-24x+144)-108+2\cdot144=-2(x-12)^2+180\)

Hence:

\(\displaystyle P_{\max}=180\)

 

[HallsofIvy]

Completing the square: -2x^2 + 48x - 108= -2(x^2- 24x)- 108.
24/2= 12 and 12^2= 144. Add and subtract 144.

-2x^2+ 48x- 108= -2(x^2- 24x+ 144- 144)- 108= -2(x^2- 24x+ 144)- 288- 108

= -2(x- 12)^2- 396.

That is a parabola opening downward. For every x, it is -396 minus something. Its maximum value, -396, occurs when x= 12.

 

[MarkFL]

Completing the square: -2x^2 + 48x - 108= -2(x^2- 24x)- 108.
24/2= 12 and 12^2= 144. Add and subtract 144.

-2x^2+ 48x- 108= -2(x^2- 24x+ 144- 144)- 108= -2(x^2- 24x+ 144)- 288- 108

= -2(x- 12)^2- 396.

That is a parabola opening downward. For every x, it is -396 minus something. Its maximum value, -396, occurs when x= 12.

The part highlighted above in red has the wrong sign. You are subtracting 288 twice.

 

[HallsofIvy]

Yes, thank you for catching that.

So the parabola is \(\displaystyle -2(x- 12)^2+ 288- 108= -2(x- 12)^2+ 180\).

That is a parabola, opening downward with vertex (12, 180). The maximum is 180 and occurs at x= 12.

Okay, that's a lot better- you have a positive​ profit!