# Math Help - 2 Optimization problem need help

1. ## 2 Optimization problem need help

Hello I need step by step all way to get the answer.

1) Find the closest distance of the point (3,1) to the curve xy=8.

2) Find the farthest distance of the point (3,0) to the curve xy=8.

2. Originally Posted by Fyou88
Hello I need step by step all way to get the answer.

1) Find the closest distance of the point (3,1) to the curve xy=8.

2) Find the farthest distance of the point (3,0) to the curve xy=8.
1) Let the point on the curve that gives closest distance have coordinates (a, 8/a).

Then $D^2 = (a - 3)^2 + \left( \frac{8}{a} - 1\right)^2 = .....$

Now use calculus in the usual way to maximise $D^2$.

2) is done in a similar way.

3. how do u set up #2?

4. Originally Posted by Fyou88
how do u set up #2?
In the same way as #1 (I thought I said that in my first reply ....)

5. so it doesnt matter if it says closest or farthest distance i would do it the same way to get the answer? what is the step after i maximize?

6. Originally Posted by Fyou88
so it doesnt matter if it says closest or farthest distance i would do it the same way to get the answer? what is the step after i maximize?
Have you been taught how to solve an optimisation problem?

You find the stationary points and then check the nature of these stationary points. If you want to maximise then you keep the solution that is a maximum. If you want to minimise then you keep the solution that is a minimum.

7. i have just learned it but don't understand too well and i'm allergic to word problems.

8. how do i do (8/x-1)^2.

9. would someone be kind enough to solve #1 with all steps listed and explain? im really lost with optimization.

10. Originally Posted by Fyou88
would someone be kind enough to solve #1 with all steps listed and explain? im really lost with optimization.
I have set it up for you! All that is essentially left is to find the coordinates of the maximum turning point of D^2. Surely you can do this? If not, you need to go back to your classnotes and textbook and thoroughly review the material that this topic is based on. To find the derivative, I suggest you apply some basic algebra and expand the expression I gave you. Then differentiate it term-by-term.

If you want more help, show some effort. We are not going to just hand you a step-by-step by solution.

11. ok. this is what i have now. and im stuck. im not sure if i did (8/x-1)^2 correct. 2x-6+64/x^2-16/x+1=0. am stuck here can someone help me out??

12. Your objective is to find the derivative of $(a-3)^{2} + \left( \frac{8}{a} - 1\right)^2$ and set it equal to 0.

Taking each part in turn:
Step1
Here is the differenciation of $f(a) = \left( \frac{8}{a} - 1\right)^2$

rewrite this as
$f(a) = \left( 8a^{-1} - 1\right)^2$

From here, just differentiate using the chain rule:
$f'(a) = 2 \left( 8a^{-1} - 1\right)^{1} * -8a^{-2}$
$f'(a) = -16a^{-2} \left( 8a^{-1} - 1\right)$

In case your chain rule is a bit rusty, it goes like this:
Bring the power of the bracket (2) down and reduce the power on the bracket by 1.

Then multiply by the deriviative of the bracket

Step2
Now, we need the derivative of:
$(a-3)^{2}$
which is $2(a-3)$

Step3
Combining those 2 results, we have the derivative we want. Your optimisation problem is:
$2(a-3) + -16a^{-2} \left( 8a^{-1} - 1\right) =0$

Have a go at that. I might have made a mistake because that equation looks pretty messy to solve.

13. so u get 2a-6-16/a^2+8/a-1=0 ??

14. Originally Posted by Fyou88
so u get 2a-6-16/a^2+8/a-1=0 ??
You are expected to know how to expand and simplify the expression I gave for D^2:

$D^2 = a^2 - 6a - \frac{16}{a} + \frac{64}{a^2} + 10$.

You are expecetd to know how to differentiate each term in the above:

$\frac{d \, D^2}{da} = 2a - 6 + \frac{16}{a^2} - \frac{128}{a^3}$.

15. $
\frac{d \, D^2}{da} = 2a - 6 + \frac{16}{a^2} - \frac{128}{a^3}
$

ok so now i set that =0? help anyone???

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