1. ## Business application calculus thread 1?

Find the number of units x that produces a maximum revenue R

1. R= ((1,000,000x))/((.02x^2+1800))

Now I think I am supposed to differentiate this then set it to zero to find the number but what would I use quotient rule?

Find the number of units x that produces the minimum average cost per unit c

C=3000x-x^2 square root(300-x)

Since cost is C/x

I did

((3000x-x^2 square root(300-x))/((x))

I am stuck here the radical is kind of confusing me if I took out the x would I just be left with

x-square root 300 ? and then I have to take the derivative of that and set it to zero?

2. ## Re: Business application calculus thread 1?

It's a good solution to calculate the first derivative and let $f'(x)=0$ and indeed use the quotientrule for the first one.

3. ## Re: Business application calculus thread 1?

Alright

I did the quotient rule to get

((2000x^2+18 eight zero+40,000x^2))/((.02x^2+1800)^2))

Now I have to solve this to find x. But I am not sure how?

Can anyone help me simplify this.

4. ## Re: Business application calculus thread 1?

There's a mistake in the numerator, it has to be:
$f'(x)=\frac{20000x^2+1800000000-40000x^2}{(0,02x^2+1800)^2}=\frac{-20000x^2+1800000000}{(0,02x^2+1800)^2}$

Solve $f'(x)=0$ and keep in my mind that a fraction becomes zero if the numerator becomes zero.

5. ## Re: Business application calculus thread 1?

What strategy could I use to solve the fraction?

6. ## Re: Business application calculus thread 1?

I gave you a hint, if you want to solve:
$\frac{-20000x^2+1800000000}{(0,02x^2+1800)^2}=0 \Leftrightarrow -20000x^2+1800000000=0$

This is a quadratic equation ...

7. ## Re: Business application calculus thread 1?

As for my second question

I have ((3000x-x^2 square root(300-x))/((x))

But I am slightly confused on simplifying it.

8. ## Re: Business application calculus thread 1?

Is the function:
$f(x)=\frac{(3000x-x^2)\cdot(\sqrt{300-x})}{x}$
?

Yes sir.

10. ## Re: Business application calculus thread 1?

Do you notice that:
$f(x)=\frac{(3000x-x^2)\cdot(\sqrt{300-x})}{x}=\frac{x(3000-x)\cdot(\sqrt{300-x})}{x}$
$=(3000-x)\cdot \sqrt{300-x}$

Use the product rule.

11. ## Re: Business application calculus thread 1?

$f(x)=\frac{(3000x-x^2)\sqrt{300-x}}{x}$

$f(x)=\frac{x(3000-x)\sqrt{300-x}}{x}$

$f(x)= (3000-x)\sqrt{300-x}$

product rule ...

12. ## Re: Business application calculus thread 1?

Applying the product rule I now have

- square root(300-x)+(3000-x)((-1))/((2 square root (300-x))=0

The question is how can I solve this

13. ## Re: Business application calculus thread 1?

Your derivative is correct:
$-\sqrt{300-x}+\frac{x-3000}{2\sqrt{300-x}}=0$

Make a common denominator ...

14. ## Re: Business application calculus thread 1?

I used 2 square root(300-x) as common denominator to get

-600+2x+x-30000=0 Did I set this up correctly?

15. ## Re: Business application calculus thread 1?

Originally Posted by homeylova223
I used 2 square root(300-x) as common denominator to get

-600+2x+x-30000=0 Did I set this up correctly?
Yes, that's correct.

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