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WebGraphing.com Forum » List all forums » Forum: Calculus Homework Help » Thread: max/mins |
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Total posts in this thread: 2 |
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strikeapose
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A rancher wishes to fence in a rectangular pen of 7200 sq yds. adjacent to a river for his cattle. What dimensions maximize the amount of fencing required? Assume he doesn't have to fence the boundary by the river. Justify your maximum using calculus. can you tell me how to do it not just give the asnwer? thanks!!! |
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pskinner
Joined: Apr 2, 2005 Posts: 797 Status: Offline |
Let x represent the fence length perpendicular to the river and y the fence length parallel to the river. The amount of fencing required is F=2x+y. At this point, F is a function of x and y. The area is then x*y=7200. I suspect you really want to "minimize" the amount of fencing required; the maximum does not exist since either x or y could be as large as one wishes. Solving the area equation for y, you get 7200/x. This can be substituted into the equation for the fencing: F=2x+7200/x=2x+7200*x^(-1) Now, F is a function of just x: F=F(x). To find any extrema, take the derivative and set it equal to zero: F'(x)=2+(-1)*7200*x^(-2)=0 Solving, you have: 2-7200/x^2=0 2=7200/x^2 2x^2=7200 x^2=3600 x=+ or - 60 Since fencing needs to be positive, x must equal 60. To have the area equal 7200, substitute 60 for x in the area equation and get y=120. If you want to use calculus to double check that when x=60 you have a minimum, take the second derivative and determine the sign when x=60: F''(x)=0+(-1)(-2)*7200*x^(-3)=14400*x^(-3) Thus, F''(60)>0 so this graph is concave up at x=60, meaning that there is a minimum at x=60. ---------------------------------------- Principal Skinner |
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Re: max/mins