"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 !

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 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\$.

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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\$.

\$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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