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?
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.
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
200=WL = 200=10(L)
so i got dd/dt to round out to about 5.367?
I too got for this problem. But i got something else for the rate of the diagonal.
This is my work:
so if the area is then must equal . But the area now is because it is a triangle. so with the formula where and and . The derivative of that would then be . Thus when we plug everything in, we get, .
to find we do which the derivative of is . when we plug everything in, we get which equals = so .
if we plug -2 back into , .
did i make any mistakes anywhere? I feel like i did, but the logic makes sense. any reactions?
Originally Posted by calc_student09