Mathematics ap24:

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? Narad:

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:

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

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