Finding the minimum in a messy equation

facepalm

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Hi, my homework problem requires that I find the minimum in this equation:

C = 100x + 60[(5000/x)-20] + 60(x-30) + 60(5000/x)

I don't know how to find the derivative of this, and I don't know how to find the minimum once I've gotten the derivative.

Please help! :shock:
 
Hi, my homework problem requires that I find the minimum in this equation:

C = 100x + 60[(5000/x)-20] + 60(x-30) + 60(5000/x)

I don't know how to find the derivative of this, and I don't know how to find the minimum once I've gotten the derivative.

Please help! :shock:

Hint:

\(\displaystyle \frac{d}{dx}[\dfrac{c}{x}] \ = \ -\dfrac{c}{x^2}\)

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
 
Hello, facepalm!

Find the minimum: .\(\displaystyle C \:=\: 100x + 60[(5000/x)-20] + 60(x-30) + 60(5000/x)\)

I don't know how to find the derivative of this,
and I don't know how to find the minimum once I've gotten the derivative.

Please help!

Do I understand you correctly?

You are doing a Calculus problem.
But you don't know how to differentiate
. . and you don't know how to use a derivative.


And evidently you don't know Algebra either,
. . otherwise you would have simplified that function.


\(\displaystyle C \;=\;100x + 60\left(\dfrac{5000}{x} - 20\right) + 60(x-30) + 60\left(\dfrac{5000}{x}\right) \)

\(\displaystyle C \;=\;100x + \dfrac{300,\!000}{x} - 1200 + 60x - 1800 + \dfrac{300,\!000}{x}\)

\(\displaystyle C \;=\;160x + \dfrac{600,\!000}{x} - 3000\)


\(\displaystyle C \;=\;160x + 600,\!000x^{-1} - 3000\)


Now review your Calculus notes or ask your teacher for help.
 
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