Lacy has \$600 set aside to build a rectangular exercise kennel for her dogs. She will buy fencing material for \$15/ft. Because the side of an existing barn will be used for one of the sides of the kennel, only three sides need to be fenced.
A.) Determine the dimensions of the kennel that will enclose the maximum area.
B.) What area will be enclosed?

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1. let the width against the barn be x ft
let the length parallel to the barn be y ft

so total length = 600/15 = 40
so she needs to fence 40 ft

2x + y = 40
y = 40-2x

area =xy
= x(40-2x)
= 40x - 2x^2

are you taking Calculus?
if so, d(area)/dx = 40 - 4x = 0 for a max of area

x = 10
then y = 20

so max area = xy = 200 ft^2, when the width is 10 ft and the length is 20 ft

if not Calculus

area = 40 - 2x^2 is a parabola opening down.
we need its vertex:
the x of the vertex is -b/(2a) = -40/-4 = 10
sub in :
area = 40(10) - 2(10^2) = 200 t^2

same conclusion

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2. My question is very similar to this one, except they say that Lacy needs the area of the kennel to be 150ft^2. So what I did was take the known information above, where she can fence 40 feet. The I figured out what times what would equal 150 and add to get 40, (15*10) So since it is along the barn wall, you can have the longer sides to equal 15 and the barn wall side to equal 10. If you add the three sides up, it will equal 40. ;)
Hope this helps!

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3. YOU GALS SO HELPFUL I GOT A 100% THANKS SOOOOO MUCH ('......')

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