My Math Forum Help with fencing question... ASAP

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 September 24th, 2019, 04:05 PM #1 Newbie   Joined: Sep 2019 From: Canada Posts: 2 Thanks: 0 Math Focus: Algebra Help with fencing question... ASAP This question from my assignment is giving me some trouble. I feel like I'm really close but my end results still end up screwed up. If anyone can help me solve this so I can hand it in tomorrow afternoon I'd definitely appreciate it ! A farmer has 120m of fencing and plans to use it to build a rectangular garden. One side of the garden will be against his house and does not require fencing. What dimensions will allow the maximum area for this garden? What is this maximum area?
 September 24th, 2019, 04:59 PM #2 Member     Joined: Oct 2018 From: USA Posts: 99 Thanks: 72 Math Focus: Algebraic Geometry $L$ is the length of the side of the fence opposite the wall of the house, $W$ is the length of the other sides. $L + 2W=120$ $A=LW$ We can solve for $L$ $L=120-2W$ So, $A=(120-2W)W$ $A=120W-2W^2$ Now we can use the vertex formula $x_v = \frac{-b}{2a}$ to find the value of $W$ where $A$ is at max. $\frac{-b}{2a} = \frac{-120}{-4} = 30 = W$ So, we know $W=30$. We use the equation for $L$, $L=120-2W$ to find $L=120-2(30) = 60$ So the area is at maximum when $L=60$ and $W=30$ Apply the Area formula to find the max area is $A=60 \times 30 = 1800$ Thanks from SimpleMinded
 September 24th, 2019, 08:42 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2271 If u and v are real and u + v is a constant, uv = ((u + v)Â² - (u - v)Â²)/4 is maximized when u = v. For the fencing problem, let u = 120 - 2W and v = 2W, then u = v implies 120 - 2W = 2W, so W = 30, etc. Thanks from greg1313 and SimpleMinded

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