Solving Revenue and Cost Function

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xxMsJojoxx

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I am not able to solve this question.
I got this.1604184206816.png
But don't know how I can solve for x. I tried factoring, but am not able to solve for x?
 
Before factoring why not divide by 5?? Can we see your factoring attempts?
Is factoring the only way you know how to solve a quadratic equation?
Have you learned about completing the square method or using the quadratic formula?
 
Jomo said:
Before factoring why not divide by 5?? Can we see your factoring attempts?
Is factoring the only way you know how to solve a quadratic equation?
Have you learned about completing the square method or using the quadratic formula?
Click to expand...
Thanks for the hint. I now tried using the quadratic formula, to find the vertex of the 2 equations.
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So, I got between 5/3 to 25.

But the correct answer is 2 to 20?
 
No, no, use the equation that you came up with, namely 5x^2 - 110x + 150. Divide this equation by 5 to get smaller numbers and then find the zeros using the method of your choice.
 
If you rely on memorizing formulas, be sure to remember what each formula is for. What you chose is not the quadratic formula. Let’s start with the formula for profit:

Profit = revenue - cost =

[MATH](- 2x^2 + 100x + 50) - (3x^2 - 10x + 200) = - 5x^2 + 110x - 150.[/MATH]
Now stop to think a moment. If x has a large absolute value, then profit will be negative because - 5x2 will be hugely negative. Thus, profit will be positive only for those values of x that lie between the x-intercepts (if there are any).

Now how do we find those x-intercepts. We set our equation to zero and use the quadratic formula.

[MATH]-5x^2 + 110x - 150 = 0 \implies x^2 - 22x + 30 = 0 \implies[/MATH]
[MATH]x = \dfrac{22 \pm \sqrt{22^2 - 4(1)(30)}}{2 * 1} = \dfrac{22 \pm \sqrt{484 - 120}}{2} = 11 \pm \sqrt{91}.[/MATH]
Now when you evaluate that you don’t get 2 or 20 as a result. So why are 2 and 20 the correct answers?
 
JeffM said:
If you rely on memorizing formulas, be sure to remember what each formula is for. What you chose is not the quadratic formula. Let’s start with the formula for profit:

Profit = revenue - cost =

[MATH](- 2x^2 + 100x + 50) - (3x^2 - 10x + 200) = - 5x^2 + 110x - 150.[/MATH]
Now stop to think a moment. If x has a large absolute value, then profit will be negative because - 5x2 will be hugely negative. Thus, profit will be positive only for those values of x that lie between the x-intercepts (if there are any).

Now how do we find those x-intercepts. We set our equation to zero and use the quadratic formula.

[MATH]-5x^2 + 110x - 150 = 0 \implies x^2 - 22x + 30 = 0 \implies[/MATH]
[MATH]x = \dfrac{22 \pm \sqrt{22^2 - 4(1)(30)}}{2 * 1} = \dfrac{22 \pm \sqrt{484 - 120}}{2} = 11 \pm \sqrt{91}.[/MATH]
Now when you evaluate that you don’t get 2 or 20 as a result. So why are 2 and 20 the correct answers?
Click to expand...
Thank you. That was the answer given in my textbook. This is a review question for a Math Course in College.
 
JeffM said:
I agree that it is the correct answer. But if you find the x-intercepts of your profit function, they are not 2 and 20. So WHY are 2 and 20 the correct answers?
Click to expand...
I don't know . I tried plugging in x in the original equation as well.

I am not understanding what the zero is calculating actually.
 

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OK

Let's go through this step by step.

Do you understand why the profit function is

[MATH]-5x^2 +110x - 150.[/MATH]
So we want to find those values for x where

[MATH]-5x^2 +110x - 150 > 0.[/MATH]
Now you drew a sketch. GREAT. But you made life a bit complicated by drawing two curves. You can just sketch the one curve for profit.

Please do that, just a crude sketch of the single curve. That is quicker and easier to understand than sketching two curves.

Your sketch should look like an upside down U.

If there are any values for which profit is positive, the curve will have to be above the x-axis. Most of the time that curve will be below the x-axis. Do you see why?

If the curve is sometimes below the x-axis and sometimes above the x-axis, it must cross the x-axis, right?

When the curve crosses the x-axis, its value is ZERO.

So the profitable region will be between those points where the function is zero. Still with me?

So the way to find the zeroes of a quadratic function (if they exist) is to use the quadratic formula.

What answer do you get when you use it?
 
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