Question

Need help for calculus!

Answer 1

Ask the question.

Answer 2

@ash2326 , I see you just got on, but I have yet to take calculus, so might you be able to assist?

Answer 3

Can you attach a screenshot on here by using attach file :/ I can't access that link.

Answer 4

@joeylee Can you write the cost of the installation?
it'll be in terms of x

Answer 5

C(x) = 10x + (26( 36+ (79-x)^2)^1/2 is this correct?

Answer 6

^ oh there shdn't be bracket before 26

Answer 7

Yes
\[C=10\times x+\sqrt{36^2+(79-x)^2}\]
Now take derivative of c with respect to x

Answer 8

I missed the 26 :(

Answer 9

\[C'(x) =\frac{10+13*2(x-79) }{ x ^{2}-158x+7537 }\]

Answer 10

^oh missed the sqrt the whole denominator

Answer 11

It should be
\[C'=10+26 \times \frac{-2(79-x)}{2\sqrt{36^2+(79-x)^2}}\]

Answer 12

Do you follow?

Answer 13

hmm.. not really,, is this following some rule?

Answer 14

\[A=x+\sqrt{2x}\]
I used this
\[A'=1+\frac{1}{2\sqrt {2x}}\times \frac{d}{dx} (2x)\]

Answer 15

yours is correct but 10 won't have anything in denominator

Answer 16

oh ya sorry about that, i kinda get where you're coming from though but i find it hard, anw, afterwards is we just equate to 0 and find x right?, and sub x to the first equation, C(x)?

Answer 17

yes, that's correct. That will give us x which minimizes cost and the minimum cost

Answer 18

thanks! :]