Relative Max

[katie9426]

I'm having trouble figuring this out. I keep getting 4 and its a multiple choice problem and 4 isn't an answer! thanks for any help!

For what value of k will 8x+k/x^2 have a relative maxium at x=4?

the choices are -32, -16, 0, 16, or 32

I used L'Hopital's Rule but I get zero on the first derivative and I didn't think that was correct.

 

[galactus]

\(\displaystyle \L\\f(x)=\frac{8x+k}{x^{2}}\)

\(\displaystyle \L\\f'(x)=\frac{-2(4x+k)}{x^{3}}\)

Now, sub x=4 into f'(x), set to 0 and solve for k.

k is definitely one of your choices.

 

[katie9426]

Two questions: a) how is the derivative x^3 and not 2x? b)when I did what you said I got 2, I did it another way and got -14, ahhh???

 

[galactus]

Use the quotient rule:

\(\displaystyle \L\\\frac{8x+k}{x^{2}}\)

\(\displaystyle \L\\\frac{x^{2}(8)-(8x+k)(2x)}{x^{4}}=\frac{-8}{x^{2}}-\frac{2k}{x^{3}}=\frac{x}{x}\cdot\frac{-8}{x^{2}}-\frac{2k}{x^{3}}=\frac{-8x-2k}{x^{3}}=\frac{-2(4x+k)}{x^{3}}\)

\(\displaystyle \L\\\frac{-2(4(4)+k)}{64}=0\)

\(\displaystyle \L\\\frac{-2(16+k)}{64}=0\)

Now, you can find k, can't you. What makes the numerator 0

 

[katie9426]

Is it -16?