Lagrange Multipliers Dimensions A farmer wishes to fence off a rectangular pasture along the bank of a river. The area of the pasture is 3200 $yd^2$. There is no fencing needed along the river bank. Find the dimensions of the pasture that'll require the least amount of fencing. $$\\\$$ ![enter image description here]( I'm not sure how to begin this problem. I know the area is $lw$ = 3200. Please help.

Since the pasture is a rectangle, the Area= $xy$= 3200.

The minimum fencing required is f($x$,$y$) = $x$+2$y$.$$\\\$$ $fx$→ 1 = λy

$\implies$ $\frac{1}{y}$ = λ

$fy$→ 2 = λx

$\implies$ $\frac{2}{x}$ = λ

Set them equal and cross multiply: $\frac{1}{y}$=$\frac{2}{x}$

$\implies$ 2$y$=$x$ $$\\\$$ Remember that the Area: $xy$= 3200 and we just found what $x$ equals.

$xy$= 3200

= (2$y$)$y$ = 3200

= 2$y^2$ = 3200

= $y^2$ = 1600

= $y$ = ±40

$y$ = 40 because you can't have a negative yard. $$\\\$$

Remember that 2$y$=$x$ and we just found what $y$ equals.

2$y$=$x$

= 2(40) = $x$

= $x$ = 80 $$\\\$$ **So your dimension are 80 x 40 yards.**