Optimizing Walkway Weight: Minimize 40x + 30y

[chompysj]

Homework Statement

You're building a walkway from the corner of one building to the corner of another building. The diagram looks like this.

The street is 100 ft wide, and 50 ft long.
The walkway will weigh 40 pounds per feet when it is parallel to the street and 30 pounds per feet when it is crossing the street.

How should the walkway be laid out to have a minimal weight?

Homework Equations

I'm going to call the horizontal portion of the walkway x ft of length, and the portion that crosses the street y ft of length. So the total length of the walkway would be x+y.
Then the equation for the total weight of the walkway would be 40x + 30y, right?
Also I'm not sure if the pythagorean theorem should come into play here.

The Attempt at a Solution

This is where I'm really, really confused. I know that when you do an optimization problem , the hard part is finding the right equation. I have no idea what that is. Can someone just give me a little prod in the right direction? Thanks.

 

[Dick]

x and y are not independent. You can write y in terms of x if you finally decide you do need the pythagorean theorem. The right triangle containing y has one leg of length 100-x and another of 50.

 

[chompysj]

So y would be:

\sqrt{100^2 + (50-x)^2}

Now do I take the derivative of that?

 

[Dick]

Use the chain rule. Besides, you actually want to take the derivative of 40x+30y, right?

 

[chompysj]

Would the derivative be

-(50-x)^{2}

But how would taking the derivative help me? Is it related to the 40 lb/ft and the 30 lb/ft?

 

[Dick]

You take the derivative and set it equal to zero. But your guess for the derivative is way off. The derivative of sqrt(f(x))=f'(x)/(2*sqrt(f(x)). How did I know that?

 

[chompysj]

Umm, all I can tell is that you must have gotten the derivative out of

(1/2) x (y') x (y^(-1/2))

But I don't really understand how you got that. See, I'm not very good at using the Chain Rule, and what I did was just

\frac{1}{2} x \sqrt{0 + (50-2)^{2}}^{-\frac{1}{2}} x (2 x (-1))

 

[Dick]

You've got a good start there but in this case y=100^2+(50-x)^2. y' isn't 2*(-1). What is it? And the denominator is sqrt(100^2+(50-x)^2), why did the the 100^2 turn to zero and the x disappear?

 

[chompysj]

I got the ( 2 * -1) part from differentiating (50-x)^2.

 

[Dick]

The derivative of (50-x)^2 is 2*(50-x)*(-1).

 

[chompysj]

Oh right!
The 100^2 turned to 0 because it is an integer and you can't differentiate it anymore. The x disappeared because it is only to the one-power.

 

[Dick]

Umm, all I can tell is that you must have gotten the derivative out of

(1/2) x (y') x (y^(-1/2))

But I don't really understand how you got that. See, I'm not very good at using the Chain Rule, and what I did was just

\frac{1}{2} x \sqrt{0 + (50-2)^{2}}^{-\frac{1}{2}} x (2 x (-1))

The sqrt comes from y^(-1/2). That's NOT (y')^(-1/2). You might want to review differentiation in general.