Maximizations and Minimizations

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Prophet

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Nov 11, 2007
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6
1. How close does the semicircle x^2 + y^2 = 16, y > 0 come to the point ((3)^(1/2), 1)? (Hint: if you minimize the square of the distance, you can avoid dealing with square roots.)

Okay, I understand that you're suppose to use the distance formula with the points ( x, [16-x^(2)]^(1/2) ) and (3^(1/2),1) and plug that into the distance formula... and then you're suppose to square both sides, I am just having trouble differentiating it.

2. The strength of a rectangular wooden beam is proportional to its width times the square of its depth. Find the dimensions of the strongest rectangular beam that can be cut from a 12-inch diameter log.

I have no clue where to start here.
 
Now I just need the first one... and no one was responding to my question on the other one :'(
 
You mean no one responded to the log question?. You better look again. :roll:
 
galactus said:
You mean no one responded to the log question?. You better look again. :roll:
Click to expand...

No no no :) I completed that question thanks to your help! the first one... :'(
 
Just like on the other forum, solve for x when you set the derivative of the "F" as you called it to 0:

\(\displaystyle F' = -2(\sqrt{3} - x) - \frac{2x(1 - \sqrt{16 - x^{2}})}{\sqrt{16 - x^{2}}\) (taken off the other forum, assuming it's correct)

F' = 0. Solve for x (looks like it's going to be rather tedious)
 
o_O said:
Just like on the other forum, solve for x when you set the derivative of the "F" as you called it to 0:

\(\displaystyle F' = -2(\sqrt{3} - x) - \frac{2x(1 - \sqrt{16 - x^{2}})}{\sqrt{16 - x^{2}}\) (taken off the other forum, assuming it's correct)

F' = 0. Solve for x (looks like it's going to be rather tedious)
Click to expand...

Haha, that's the problem... I am having trouble setting it equal to 0!!! :'(
 
What do you mean?

\(\displaystyle 0 = -2(\sqrt{3} - x) - \frac{2x(1-\sqrt{16-x^{2}})}{\sqrt{16-x^{2}}\)

You mean you don't know how to solve for x? Show us what you've done and maybe we can help you out.
 
o_O said:
What do you mean?

\(\displaystyle 0 = -2(\sqrt{3} - x) - \frac{2x(1-\sqrt{16-x^{2}})}{\sqrt{16-x^{2}}\)

You mean you don't know how to solve for x? Show us what you've done and maybe we can help you out.
Click to expand...

I've got down to x=square root of 3, annnnddd this long one, 2(3^1/2) - 1 = (-) [1 - (16-(x^2))]/ ( 16- x^2)^1/2
 
Hmm don't know what you mean there and I don't think \(\displaystyle \sqrt{3}\) is the answer. Can you use your graphing calculator to solve for it? It seems really harsh to go through all the algebra involved there.

Oh, and looks like the derivative calculated has a minor mistake. It should be:

\(\displaystyle F' = -2(\sqrt{3}-x) + \frac{2x(1-\sqrt{16-x^{2}})}{\sqrt{16-x^{2}}\)

There's an addition sign instead of a subtraction sign when you do the derivative of the two expressions of F.

And from my graphing calculator, x is about 3.4641027.
 
\(\displaystyle \L\\D^{2}=(x-\sqrt{3})^{2}+(\sqrt{16-x^{2}}-1)^{2}\)

=\(\displaystyle \L\\-2\sqrt{16-x^{2}}-2\sqrt{3}x+20\)

Now, take the derivative:

\(\displaystyle \L\\\frac{2x}{\sqrt{16-x^{2}}}-2\sqrt{3}\)


Now, set to 0 and solve for x. The solution is rather obvious now.

You can finish now?. Just simple algebra. I am sorry to say, if you are havng trouble solving this, then perhaps calculus wasn't the best choice now.
 
Wow that's much simpler to deal with. :oops:

Sorry for the misguidance. Was going with what the other forum had!
 
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