Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

Calculus-Optimization: The product of two positive numbers is 54. Fnd the numbers if the sum f the first number plus the square of the second number is as small as possible

OpenStudy (anonymous):

malevolence u disappeared again lol your just giving me the answer and i dont understand

OpenStudy (anonymous):

Okay from x+(54^2/x^2)=s. Taking the derivative. What confuses you?

OpenStudy (anonymous):

how u found the derivative of the 2nd part...

OpenStudy (anonymous):

Okay. (i'm going to call 54^2=a, its easier to write) Rewrite it as: x+ax^(-2)=s. Because a/x^2=ax^(-2). Follow me up to that point?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

Okay, then you have a power rule d/dx(x^n)=nx^(n-1) So: It becomes 1+(-2)a(x^(-3))=0. Right?

OpenStudy (anonymous):

ok so we just pretty much ignore the 54...thats what confused me i thought since it was (54/x)^2 we would use the chain rule...no?

OpenStudy (anonymous):

The 54 is just a constant, you can pull it out of a derivative. So its immaterial. (I just called it a for convenience).

OpenStudy (anonymous):

You can use the chain rule. You would have (2)(54x^-2)(-1)(54/x)

OpenStudy (anonymous):

Convince yourself that those are both equally valid. :P

OpenStudy (anonymous):

ok...i'm sorry for my slow moment im very good with derivatives but...with optimization i dont know if i should multiply or which rule's i should use thanks so much lol

OpenStudy (anonymous):

u think if i use the chain rule i would get the same answer?

OpenStudy (anonymous):

Thats what I did above^^ Yes you will. If you multiply what I have together up there, you get the same thing.

OpenStudy (anonymous):

oh ok THANKS SO MUCH..i dont know why im so confused with optimization

OpenStudy (anonymous):

No problem, after you do it a bunch it'll become easy. :P Just post any questions you have and I'll try to explain a little better.

OpenStudy (anonymous):

ok! thanks

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!