Calculus: Wall/fence problem, minimizing cost

[Femme_physics]

Homework Statement

A rectangle site, ABCD, whose area is 4500 meter squared, is adjacent to a wall on one side (see drawing).

http://img820.imageshack.us/img820/4457/qcalc.jpg


They fence the front of the site, BC, and its sides, AB and CD. The price of installing a fence at the front of the site (BC) is 16 shekels per meter, and the price of installing a fence on the sides (AB and CD) is 10 shekels a meter. What should be the length of the front of the site, so the price of installig the fence would be minimal?

Homework Equations

Calculus...

The Attempt at a Solution

attached. I basically tried to make a formula that relates area, prices and side lengths, and then took its derivative, set it equal to zero, find a min point (found only 1 point...) and then use the score at the original formula...

here's my attempt...how far off am I?

 

[issacnewton]

Shalom femme

I think your area formula is wrong. Area is just WL.

then price function would be f = 16L+10W+10W

But WL = 4500

 

[Femme_physics]

Shalom gam leha :)

Yes area is WL, but each W and L each come with their own price, so I figured it makes sense. No?

Otherwise, can you hint me at the right direction, please?

 

[issacnewton]

I have already written the price function. Since WL =4500 , W= 4500/L. so the price
function is

$$f(L) = 16L+ 20\left(\frac{4500}{L}\right)$$

and you have to minimize this price function...

 

[Femme_physics]

Oh, did you edit your post later? I only read the 1st two lines. Sorry. And thanks.

I'll go back trying to solve it now :)

 

[Femme_physics]

I'm getting imaginary answers when I set the derivative equal to 0...that can't be it right?

 

[jhae2.718]

Don't bother with a common denominator.
$$a-\frac{b}{x^2}=0 \Rightarrow a = \frac{b}{x^2} \Rightarrow ax^2 = b$$
with the condition that $x \neq 0$.

Also, from lower on in your work, $$\frac{16L^2-90000}{L^2} = 0$$ is not 16L²-90000=L²

 

[issacnewton]

femme, you are getting

$$16L^2 = 90000$$

its simple equation. check your calculations again. no imaginary numbers here.

 

[Femme_physics]

Excellent :) 75, and -75. Making perfect sense now.

And sorry for my algebraic mishap. I'm still knocking myself on the basics. You guys are great :)

 

[issacnewton]

ok... on physical grounds we reject -75 ...

 

[jhae2.718]

Also, regarding the algebra, don't always head straight for the quadratic formula. Sometimes there are far easier methods of solution.

It's also good to verify that your answer is a minimum by taking the second derivative. (It would be bad if, in a case of multiple solutions, we picked one that maximized cost! )

 

[Femme_physics]

@ Isaac - Yes, I realize -75 doesn' make sense based on physical grounds. Thank you :)

@ jhae - I checked it's a min by setting a number lower than 75 in the derivative and seeing a negative figure, then a number over 75, and seeing a positive figure. Definitely a min. Thanks :)

Also, regarding the algebra, don't always head straight for the quadratic formula. Sometimes there are far easier methods of solution.

Actually using a calculator to solve a quadradic formula, so it's pretty easy (I just plug in a, b, and c)

 

[jhae2.718]

Actually using a calculator to solve a quadradic formula...

 

[Femme_physics]

LOL, why the shocked face?

 

[jhae2.718]

I'm generally of the opinion that calculator equation solvers are a black box to which most people plug numbers in and then "The Answer" comes out. (I'm not saying you're one of these, but just relating my general experience.) I think that you should only be allowed to use solvers that you have programmed yourself. (Well, for trivial cases, at least.)

Of course, I'm a hypocrite who uses MATLAB for everything, so...

 

[Femme_physics]

Heh, well, I guess it's good to practice the mind, I'm also in engineering/physics so the general mood is "don't waste too much time on math because there's so much to learn, just plug it in if you can".