# Optimization Problem with Triangles

• May 8th 2011, 07:14 PM
RezMan
Optimization Problem with Triangles
hey guys can you help me here.

A right triangle has area 50.
a) Express the hypotenuse as a function of a single variable, and
b) Find the shortest possible hypotenuse this triangle can have

I know that bh/2=50. They're asking me to write b^2+h^2=c^2 in terms of a single variable?
• May 8th 2011, 07:28 PM
TheChaz
Quote:

Originally Posted by RezMan
hey guys can you help me here.

A right triangle has area 50.
a) Express the hypotenuse as a function of a single variable, and
b) Find the shortest possible hypotenuse this triangle can have

I know that bh/2=50. They're asking me to write b^2+h^2=c^2 in terms of a single variable?

You're almost there:
bh/2 = 50 so
bh = 100
b = h/100 [MODERATOR EDIT]: See below for correction.
We want c = sqrt( b^2 + h^2 ), which becomes
c = sqrt( (h/100)^2 + h^2 )
You can rewrite that as needed.

b) Find c'.
Set it = 0.
etc
• May 8th 2011, 07:37 PM
RezMan
wow you make it seem so simple! I was wondering, doesn't sqrt(b^2+h^2) turn into just b+h. And isn't it 100/h not h/100. I can't thank you enough for your help!
• May 8th 2011, 07:46 PM
mr fantastic
Quote:

Originally Posted by RezMan
wow you make it seem so simple! I was wondering, doesn't sqrt(b^2+h^2) turn into just b+h. Mr F says: And I suppose you also want to turn sqrt(1^2 + 1^2) into 1 + 1 ....

And isn't it 100/h not h/100. Mr F says: Correct.

I can't thank you enough for your help!

..
• May 8th 2011, 07:57 PM
RezMan
Quote:

Originally Posted by mr fantastic
And I suppose you also want to turn sqrt(1^2 + 1^2) into 1 + 1

I guess not?
so what would I do with $\sqrt{\frac{10000}{h^2}+h^2}$? Should I take the derivative of that or could I reduce it more
• May 8th 2011, 08:06 PM
TheChaz
Quote:

Originally Posted by RezMan
I guess not?
so what would I do with $\sqrt{\frac{10000}{h^2}+h^2}$? Should I take the derivative of that or could I reduce it more

Yeah, maybe if someone would be so kind as to edit my mistake above...! Glad you could see the intended result...

There may be a nice trig substitution, but you can always just crank out the derivative...
• May 8th 2011, 08:22 PM
RezMan
I feel this is wrong but I don't know what else to do. ANyway. so Quotient rule?

$\sqrt{\frac{h^2(0)-10000(2h)}{h^4}+h^2 }$
• May 8th 2011, 08:35 PM
RezMan
I'm gonna have to put this one on hold. I emailed my teacher for help. Thanks a lot!
• May 10th 2011, 07:13 PM
TheChaz
Quote:

Originally Posted by RezMan
I guess not?
so what would I do with $\sqrt{\frac{10000}{h^2}+h^2}$? Should I take the derivative of that or could I reduce it more

This equals
$( \frac{10000}{h^2}+h^2)^{\frac{1}{2}}$

We take the derivative, first using the power rule, then the chain rule, then... more rules.
The derivative is

$\frac{1}{2} * (\frac{10000}{h^2}+h^2)^{\frac{-1}{2}}$...

times the derivative of
$\frac{10000}{h^2}+h^2}$ ... etc.
• May 10th 2011, 07:16 PM
mr fantastic
Quote:

Originally Posted by RezMan
I have no idea how to solve this (Angry)(Worried)(Sweating)

You have the rule for the hypotenuse H in terms of sidelength h. You are expected to find dH/dh using the chain rule, equate to zero etc. etc. Surely you have similar examples in your class notes or textbook.
• May 12th 2011, 09:54 PM
RezMan
I had a hard time finding the derivative for this. I used a website that said the derivative was this: $\frac{h^4-10000}{h^2\sqrt{h^4+10000}}$

But I only got this far: $\frac{h^4-10000}{h^3\sqrt{\frac{10000+h^4}{h^2}}}$

can someone please tell me if how to get to the one the website gave.
• May 12th 2011, 10:28 PM
mr fantastic
Quote:

Originally Posted by RezMan
I had a hard time finding the derivative for this. I used a website that said the derivative was this: $\frac{h^4-10000}{h^2\sqrt{h^4+10000}}$

But I only got this far: $\frac{h^4-10000}{h^3\sqrt{\frac{10000+h^4}{h^2}}}$

can someone please tell me if how to get to the one the website gave.

In calculus you are expected to be competent with basic algebra. It is essential. Otherwise you are in for a world of pain. It's often been said that most problems students have in calculus are due to poor skills in algebra. I suggest you extensively review all the algebra that is pre-requisite for the subject you are studying.

For the particular problem you're working on, you're meant to note that $\sqrt{\frac{10000 + h^4}{h^2}} = \frac{\sqrt{10000 + h^4}}{h}$ using a basic surd rule .... Now simplify your answer.
• May 13th 2011, 05:16 PM
RezMan
h= $\pm \sqrt[4]{10000}$
$\sqrt{\frac{10000}{h^2}+h^2}$ ?