A farmer plans to use 21 m of fencing to enclose a rectangular pen having an area of 55 m². Only three sides of the pen need fencing because of an existing wall. Find the dimensions of the pen.
For this problem you need to deal with both perimeter and area so let's start with those formulas If the length is denoted by l and the width by w: P = 2l + 2w A = l Now we only are concerned with three sides of the pen when dealing with perimeter so using the 21m of material he has, we see that 21 = l + 2w (since the other length is using the existing wall. We want the area to be 55m^2 so plugging this in we have 55 = lw Solving the perimeter equation for l and substituting, we get\[55=lw=(21-2w)w=21w-2w^{2}\]\[-2w^{2}+21w=55\] We can now solve this quadratic by moving the 55 to the left and factoring \[-2w^{2}+21w-55=0\]\[2w^{2}-21w+55=0\](2w - 11)(w - 5)=0 Solving each of these factors for w will give us two posible values for the width which we can plug into the perimeter equation to get the associated lengths. I leave the rest up to you...
5 by 11 (by eye)
That's only one of the answers @estudier. Since it's a quadratic, there are 2 solutions.
5.5 m by 10 m or 5 m by 11 m
The other is probably something daft, like a negative (guessing)
thats what i got
What's the max area you can enclose?
@nany - You are correct! Outstanding work! @estudier - No, the other solution is not daft. Many of these area/perimeter problems have to viable solutions. One is usually nice numbers like 5 x 11 and the other is less nice because you're reallocating "blocks" from the side and redistribuiing them to the length. Perhaps, @estudier, you should work out the problem using the corret method prior to assuming answers and giving false information.
Thank You So Much For Your Help !!
A = x(21-2x) so dA = 21-4x = 0 for max A --> x=21/4 and max A = 55 and an eighth. @mtbender74, I gave no false information (My first answer I said was by eye and the second I said was a guess, I apologize if you misunderstood me).
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