What is the initial step?

Question

A three-sided fence is to be built next to a straight section of river, which forms the fourth side of a rectangular region, as shown in the diagram below. The enclosed area is to equal $1800 m^2$ and the fence running parallel to the river must be set back at least $20 m$ from the river. Determine the minimum perimeter of such an enclosure and the dimensions of the corresponding enclosure.

I am using this for the first part: $1800m^2 = l(w- 20)$

Every time I use this I can't really figure out how to differentiate, which leads me to believe I'm doing something wrong. I know that one of the sides needs to be subtracted by $20$, but I can't understand when it should be subtracted.

Also, if I do find the new width, would that change the entire area?

Answer 1

We have $$1800 = l \cdot w$$

where $$l \ge 20$$

We want to minimize $2l+w$.

The Lagrangian is given by $$\mathcal{L}(l)= 2l + 1800/l +\lambda(20-l)$$

Then, $$\frac{d}{dl}\mathcal{L}(l)= 2 - l^{-2} 1800 -\lambda$$

We conclude that

$$\frac{1}{l}=\sqrt{\frac{2-\lambda}{1800}}$$

Therefore, if $\lambda=0$,

$$\frac{1}{l}=\sqrt{\frac{2}{1800}}$$

then $l=30$, which is the solution.

Candidate solutions: $l=20$ and $w=90$ or $l=30$ and $w=60$.

Solution: $l=30$ and $w=60$.

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Alternative problem: We want to minimize $2(l+w)$.

The Lagrangian is given by $$\mathcal{L}(l)= 2(l + 1800/l) +\lambda(20-l)$$

Then, $$\frac{d}{dl}\mathcal{L}(l)= 2(1 - l^{-2} 1800) -\lambda$$

We conclude that

$$\frac{1}{l}=\sqrt{\frac{1-\lambda/2}{1800}}$$

Therefore, either $\lambda=0$, implying that $l = w = \sqrt{1800}$ or $l=20$ and $w=90$.

Solution: $l=w=\sqrt{1800}$.

Answer 2

Let the lenght and breadth of the rectangle $x,y$ respectively. Then the area of the rectangle is:

$$\mathcal A=xy=1800$$ or $y=1800/x$ with $x\ne 0$. Also, it is know the perimeter is the sum of all sides, or

$$p(x)=2x+y=2x+\frac{1800}{x}$$

Differentiating the function $p(x)$ we have:

$$p'(x)=2-\frac{1800}{x^2}$$ Putting this $p'(x)=0$ we can compute the critical points. We get $x=30$ as a critical point. We see that when $x>30$, $p'(x)>0$. Also when $0<x<30$, $p'(x)<0$. By first derivate test we see that $x=30$ is the point of minimum. Thus $y=1800/30=60$. Therefore, the dimensions are $30 \times 60$ and the perimeter is $120$ feets.