Question

Hey can u guys help me with this question?Am totally stuck with it and will be really grateful if u can help! Qn attached at comment!:)

Answer 1

HERE IT IS! SORRY FOR THE LATE ATTACHMENT! HOPE U CAN HELP:)

Answer 2

Interesting, this looks like something that might involve lagrange multipliers?
This is vector calculus, right?

Answer 3

We'd say here that:\[
g(x,y)=128x^2-16x^2y+1=0 \\
f(x,y) = x+y
\]

Answer 4

I'm a bit rusty on these, but I'm referencing
http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx

Answer 5

\[\nabla f=\lambda \nabla g\]Meaning:
\[
f_x = \lambda g_x\\
f_y=\lambda g_y
\]

Answer 6

Can you take it form here?

Answer 7

The method that u used is unfamiliar to me. Thanks for ur effort anyway!:)

Answer 8

What method are you familiar with?

Answer 9

Maybe this is the method you're supposed to learn?

Answer 10

I'm not sure. I'll try to understand it through the tutorial then!

Answer 11

If you need help, I can try to work it out with you. Like I said, I'm rusty. This method might not even work well for this case.

Answer 12

After looking at the tutorial just now, I'm sure that I've not learn anything about lagrange.. I'd there any other methods beside this?

Answer 13

Nothing methodical unfortunately, no.

Answer 14

Is it calculus class?

Answer 15

Since you're limited to the positive numbers, I would say that
\[128x^2 - 16x^2y + 1 = 0\]
gives
\[y = \frac{128x^2 + 1}{16x^2}\]
Now if we let x + y = k, y = k - x so that
\[k = \frac{128x^2 + 1}{16x^2} + x\]

Now I think you can just find the minimum on \((0, \infty)\), by using a derivative

Answer 16

Thankyou understood:)