+37170 Let x= the perpindicular sides total fence needed = 5x
the remaining LONG side of the fence (opposite the building) will then be 210 - 5x
the area will then be x (210-5x) = 210x -5x^2 = area
Take derivative and set to zero to find 'x' that makes biggest area:
210 -10x = 0 10x= 210 x= 21 5x = 105 Long side = 105
Total area will be 21 x 105 = 2205 sq ft and each area will be 551.25 sq ft
Sorry! Made a math error there!
+130516 Let x be the longer side of the rectangle and let y be the shorter side of the rectangle
And the expression for the total fencing used is
210 = x + 5y solving for y, we have
y = [ 210 - x] / 5 this will be one side of a kennel
The other side will = the long side divided by 4 = x / 4
So....the area of one kennel =
[x /4] * [210 -x]/ 5 simplify
[210x - x^2] / 20 =
(-1/20)x^2 + 10.5x
We can find the x that maximizes this by -b/2a where b = 10.5 and a = -1/20
So we have
-10.5 / [ 2 (-1/20)] = 10.5 / (1/10) = 105 ft ....and this is the length of the longest side
So the length of the top side of one kennel = 1/4 of this = 105/4 ft = 26.25 ft
And the length of the other side of one kennel = [210 - 105] / 5 = 105/5 = 21 ft
So.....the dimensions that maximize the area of each kennel = 26.25 ft x 21 ft
