Solve the problems below that pertain to constructing a fence. A farmer wishes t

Question

Solve the problems below that pertain to constructing a fence. A farmer wishes to enclose 12,100 ft2 of land along a river by three sides of fence (the river forms the fourth side of the rectangular area). Find the dimensions which require the minimum length of fence. (Let x be the length of the sides perpendicular to the river, and y the length of the parallel side.) x= y = How does the answer from part (a) change if the fence is divided into three equal sections by two additional lengths of fence parallel to the river. x = y = Suppose that, in the situation from part (b), the farmer is not limited by the area of the fence, but instead has only 1040 feet of fence available. What is the largest area he can enclose? ft2

Explanation / Answer

we wnt to mimimixe z=2x+y now xy=12100 so y=12100/x hence z=2x+12100/x= now dz/dx=2-12100/x^2 for dz/dx=0 we have 2-12100/x^2=0 hence x=110/sqrt2 and -110/sqrt2 now d(dz/dx)/dx=12100/x^3 so we get minimum for x=110/sqrt2 y=12100/x=110sqrt2 b) now we need to minimize z=2x+3y where 3xy=12100 so y=12100/3x so z=2x+12100/3x hence dz/dx=2-12100/3x^2 for dz/dx=0 we have x=110/sqrt6 and y=12100/3x=110sqrt6/3=110sqrt2/sqrt3 so both x and y gets divided by sqrt 3 c)z=1040=2x+3y=>x=(1040-3y)/2=520-1.5y now we need to maximize 3xy since 3 is constant we need to maximize xy let p=xy then dp/dy=x+ydx/dy=x+y(-1.5) for dp/dy=0 we have x=1.5y 520-1.5y=1.5y 520=3y y=520/3 hence x=520-260=260

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