Optimization on area , rectangle with fixed length on 3 sides.
By Sophia Vance
I've stumbled with the problem below
"Some unused land is adjacent to a straight canal. A gardener wants to use 200 meters of fence in order to create a rectangular garden, by using the fence for three sides and the canal as the fourth. For which dimensions is the area of the garden as large as possible? "
I don't quite seem to create a general idea of what to do, what really disorients me is the fact that we have a fixed 200m length on 3 sides. I tried to set up the $b*h$ equation for the area but when trying to optimize it I get redundant with every step , I believe I don't have the correct approach to the problem..
Any kind of help or guidance will be highly appreciated ! Thank you !
$\endgroup$3 Answers
$\begingroup$Yes Joel, as you say in your comment to Sanath's answer, the constraint is not $2(x+y)=200$ but rather $2x+y=200$. Formally the problem is stated as:
$$\max \:\: xy \:\:\:\:s.t. \:\:\:2x+y=200$$
The Lagrangian is:
$$L=xy-\lambda(2x+y-200)$$
First order conditions:
$$\frac{\partial L}{\partial x}=y-2\lambda=0$$
$$\frac{\partial L}{\partial y}=x-\lambda=0$$
$$\frac{\partial L}{\partial \lambda}=2x+y-200=0$$
Solving these equations gives you:
$$x=50$$ $$y=100$$
The "shadow price" or marginal value of additional area is $\lambda=50$
$\endgroup$ 0 $\begingroup$The area of the rectangle is $l*w$, and we have the constraint that $2w+l=200$. This gives us $l=200-2w$, and the area of the rectangle is $w(200-2w)=-2w^2+200w$.
Differentiate this equation and find the critical points.
$$-2w^2+200w\to-4w+200$$
There is a critical point at $w=50$. By the second derivative test, you can also see that this point is a maximum. Thus your area is maximized when $w=50$ and $l=100$.
$\endgroup$ $\begingroup$$2l+b=200$. Thus, $b=2(100-l)$. Hence, $A=lb=200l-2l^2$. Hence, $A^\prime=200-4l=0\implies l=50,b=100$.
The rectangle with maximum area with fixed finite perimeter is a square. (Exercise.)
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