# Optimizing a rectangle

• Mar 16th 2009, 06:17 PM
calc_student09
Optimizing a rectangle
A rectangle has a constant area of 200 square meters and its length L is increasing at the rate of 4 meters per second
1)Find the width W at the instant the width is decreasing at the rate of .5 meters per second.
2) At what rate is the diagonal D of the rectangle changing at the instant when the width W is 10 meters?

1)So I know that DL/Dt= 4 and Dw/dt= -.5
So the equation is A=LW, the derivative being Da/dt= L(dw/dt)+W(dL/dt)
Would that be 0=L(-.5)+W(4)
What would I do next there?

2)How would I start this one off?
• Mar 16th 2009, 06:30 PM
wytiaz
1) excellent start. now replace either w with (200/l) or replace l with (200/w) in your derivative equation, then solve for the one remaining variable.

2) the length of the diagonal, if you forget the formula, can easily be rediscovered by using a theorem by a fellow named pythagoras. go ahead, draw a diagonal in and find it's length in terms of l and w.
• Mar 17th 2009, 06:23 PM
calc_student09
Wow, thanks, I cant believe I forgot that the area was 200. So when I sub that in I get w=5.
For the diagonal the equation would be L^2 + W^2 = D^2
2L(dl/dt)+2w(dw/dt)=2D(dd/dt)
200=WL = 200=10(L)
L=20

20^2+10^2=D^2
D=√500

A=LW
0=20dw/dt+10dL/dt
-20dw/dt=10(4)
dw/dt=-2
2(20)(4)+2(10)(-2)=√500(dd/dt)
so i got dd/dt to round out to about 5.367?
• Mar 18th 2009, 06:44 PM
OnMyWayToBeAMathProffesor
I too got $w=5$ for this problem. But i got something else for the rate of the diagonal.

This is my work:

$w=10$ so if the area is $200m^2$ then $L$ must equal $20$. But the area now is $100m^2$ because it is a triangle. so with the formula $x^2+y^2=d^2$ where $x=width=10$ and $y=height=20$ and $d=hypotenuse=\sqrt{500}$. The derivative of that would then be $2x\frac{dx}{dt}+2y\frac{dy}{dt}=2d{dd}{dt}$. Thus when we plug everything in, we get, $2(10)\frac{dx}{dt}+2(20)\frac{4}{1}=2\sqrt{500}\fr ac{dd}{dt}$.

to find $\frac{dx}{dt}$ we do $A=\frac{1}{2}xy$ which the derivative of is $A'=\frac{1}{2}x\frac{dy}{dt}+\frac{1}{2}y\frac{dx} {dt}$. when we plug everything in, we get $0=\frac{1}{2}10\frac{4}{1}+\frac{1}{2}20\frac{dx}{ dt}$ which equals $0=20+10\frac{dx}{dt}$ = $-20=10\frac{dx}{dt}$ so $\frac{dx}{dt}=-2$.

if we plug -2 back into $2(10)\frac{dx}{dt}+2(20)\frac{4}{1}=2\sqrt{500}\fr ac{dd}{dt}$, $\frac{dd}{dt}=2.683$.

did i make any mistakes anywhere? I feel like i did, but the logic makes sense. any reactions?
• Mar 18th 2009, 06:55 PM
skeeter
Quote:

Originally Posted by calc_student09
A rectangle has a constant area of 200 square meters and its length L is increasing at the rate of 4 meters per second
1)Find the width W at the instant the width is decreasing at the rate of .5 meters per second.
2) At what rate is the diagonal D of the rectangle changing at the instant when the width W is 10 meters?

1)So I know that DL/Dt= 4 and Dw/dt= -.5
So the equation is A=LW, the derivative being Da/dt= L(dw/dt)+W(dL/dt)
Would that be 0=L(-.5)+W(4)
What would I do next there?

you know LW = 200 ...

2)How would I start this one off?

how about D^2 = L^2 + W^2 ?

.