Answer:
Total area: 19802.5 ft2
Step-by-step explanation:
The total fencing being used will be the perimeter of the rectangular area plus 3 sides that will divide the total area in 4 pens. These sides will be parallel to the width of the total area, as the width will be the smaller side of the rectangle.
Calling the length L and the width W, we have that:
2L + 2W + 3W = 890
2L + 5W = 890
The total area will be:
A = L*W
From the first equation, we have:
L = 445 - 2.5W
Using this in the area equation, we have:
A = (445 - 2.5W)*W = 445W - 2.5W^2
To find the width that gives the maximum area, we find the vertix of the quadratic equation:
W_vertix = -b/2a
Where a and b are coefficients of the quadratic equation (in our case, a = -2.5 and b = 445). So:
W_vertix = -445/(-5) = 89
The maximum area will be:
A = (445 - 2.5W)*W = (445 - 222.5)*89 = 19802.5 ft2
Answer:
19802.5 ft²
Step-by-step explanation:
Perimeter given is 890ft. According to the figure(see attachment) ,
the equation will be
2x + 5y = 890
5y = 890 - 2x
y= 178- (2/5)x -->eq(1)
As we know that Area 'A' = xy
Substituting the value of 'y' in above equation
A= x . [178- (2/5)x]
A= 178x - [tex]\frac{2}{5}[/tex] x²
Next is to derive the equation
A' = - [tex]\frac{4}{5}[/tex]x + 178
Equating A' to zero in order to find the critical numbers to get your potential minimums & maximums.
- [tex]\frac{4}{5}[/tex]x + 178 =0
- [tex]\frac{4}{5}[/tex]x = -178
x = 178/ 0.8
x= 222.5
plugging the above value in eq(1)
(1)=>y= 178- (2/5)(222.5)
y= 178- 89
y= 89
Therefore,
A= x.y => (222.5) (89) => 19802.5 ft²
The largest possible total area of the four pens is 19802.5 ft²
