- **Algebra**
(*http://mymathforum.com/algebra/*)

- - **Find the dimensions of the rectangular corral split into two pens ...................**
(*http://mymathforum.com/algebra/343028-find-dimensions-rectangular-corral-split-into-two-pens.html*)

Find the dimensions of the rectangular corral split into two pens ...................1 Attachment(s) Hello sorry if I already posted this one? I did check threads I have started but did not see it.... Find the dimensions of the rectangular corral split into two pens (sharing a fence) of the same size producing the greatest enclosed area given 630 feet of fencing. I know it's not really much that I have done or remember but I think this was my first step? Which was also confusing for my tutors and teachers to at first figure out until I was able to tell them that you have 3 widths now, and four lengths. (Even the word "pens" was confusing at first for the language barrier ) thanks for your help I will do my best to grasp what your're able to show me. |

$630 = 4L + 3W$ $L = \dfrac{630-3W}{4}$ $A = 2LW = 2 \dfrac{630-3W}{4}W$ $A = \dfrac 1 2(630W - 3W^2) = -\dfrac 3 2 (W^2-210W)$ $A = -\dfrac 3 2 \left((W-105)^2 -(105)^2\right)$ $A = -\dfrac 3 2 (W-105)^2 + \dfrac 3 2 (105)^2$ The left term is negative so making it zero will maximize the area. This clearly occurs at $W=105$ $L = \dfrac{630-315}{4} = \dfrac{315}{4} =78.75$ thanks to Skipjack |

You are given that 4L + 3W = 630 ft. To maximize 2LW, you need 4L = 3W, so W = 105 ft and L = (315/4) ft. (Note: romsek calculated (630 - 420)/4, but should have calculated (630 - 315)/4.) |

All times are GMT -8. The time now is 05:25 PM. |

Copyright © 2019 My Math Forum. All rights reserved.