A rancher wants to fence in an area of 3000000 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use?

Question

Narad · Answer

Let length of field= l  and width of field=w

Area, A = lw

Length of fence= P=2l+3w

$$P=2l+\frac{ 3A }{ l }$$

$$\frac{ dP }{ dl }= ........$$

Then set $$\frac{ dP }{ dl }=0$$

and you'll get the minimum length........

surjithayer · Answer

$$\frac{ dP }{ dl }=2+\frac{ -3A }{ l^2}$$
$$\frac{ dP }{ dl }=0~gives$$
$$2-\frac{ 3A }{ l^2 }=0$$
$$2l^2=3A$$
$$2l^2=3\times3000000$$
$$2l^2=9000000$$
$$l^2=4500000$$
$$l=\sqrt{4500000}=1500\sqrt{2}$$
$$\frac{ d^2P }{ dl^2 }=0+\frac{ 6A }{ l^3 }>0 ~for~l=1500\sqrt{2}~ft$$
so P is minimum at this value.
$$w=\frac{ 3000000 }{ 1500\sqrt{2}}=\frac{ 2000 }{ \sqrt{2} }=1000\sqrt{2}$$
$$shortest~length~P=2l+3w=2\times 1500\sqrt{2}+3\times 1000\sqrt{2}=6000\sqrt{2}~ft.$$
$$\approx 8485.2814~ft$$