"Some unused soil is adjacent to a right canal. A gardener desires to usage 200 meters of fence in order to create a rectangle-shaped garden, by utilizing the fence for 3 sides and also the canal together the fourth. For which size is the area the the garden as big as possible? "

I don"t rather seem to produce a general idea of what to do, what really disorients me is the fact that we have actually a addressed 200m length on 3 sides. Ns tried to collection up the $b*h$ equation because that the area but when trying to optimize that I get redundant v every action , I think I don"t have the correct method to the problem..

Any kind of assist or guidance will certainly be very appreciated ! thank you !


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Yes Joel, together you to speak in your comment to Sanath"s answer, the constraint is not $2(x+y)=200$ but rather $2x+y=200$. Officially the problem is proclaimed as:

$$max :: xy ::::s.t. :::2x+y=200$$

The Lagrangian is:

$$L=xy-lambda(2x+y-200)$$

First bespeak conditions:

$$fracpartial Lpartial x=y-2lambda=0$$

$$fracpartial Lpartial y=x-lambda=0$$

$$fracpartial Lpartial lambda=2x+y-200=0$$

Solving this equations gives you:

$$x=50$$$$y=100$$

The "shadow price" or marginal value of additional area is $lambda=50$


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The area the the rectangle is $l*w$, and also we have actually the constraint the $2w+l=200$. This gives us $l=200-2w$, and the area of the rectangle is $w(200-2w)=-2w^2+200w$.

You are watching: Maximum area of a rectangle with 3 sides

Differentiate this equation and find the vital points.

$$-2w^2+200w o-4w+200$$

There is a an important point at $w=50$. By the 2nd derivative test, friend can likewise see that this point is a maximum. Hence your area is maximized once $w=50$ and also $l=100$.


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$2l+b=200$. Thus, $b=2(100-l)$. Hence, $A=lb=200l-2l^2$. Hence, $A^prime=200-4l=0implies l=50,b=100$.

The rectangle with maximum area with resolved finite perimeter is a square. (Exercise.)


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