Answer:
Max.area = L^{2} /8 ftx^{2}
Dimensions of maximum pen are:
x = L/2 ft (the largest side of the rectangle
y = L/4 ft
L is the length of flexible fencing material
Step-by-step explanation:
The straight brick wall with the fencing flexible material will form a rectangle of sides "x" and "y"
So we will use two concepts:
1.Area of rectangle : A=x×y
2.The length of the flexible fencing material L
According to problem statement
L = y + x + y or L = 2y + x 2y = L - x and y =(L-x)/2 (1)
And as Area of the pen is f(a) = x×y (2)
Replacing in equation (2) "y" by its value from equation (1) we have f(a) as function of one variable "x"
f(x) = x×( L- x )/2
We should get the derivative of f(a) and call it f´(a)
f¨(x) = 1 × (L - x)/2 +( x × (-1/2)) or f¨(x) = ( L - x )/2 - x/2
The second derivative f¨¨(a) = - 1/2 (negative meaning a maximun for the function)
Equating f¨(a) = 0
f¨(x) = ( L - x )/2 - x/2 = 0
L/2 - x/2 - x/2 = 0
L/2 - x = 0
x = L/2 (the largest side f the rectangle)
To get "y" (the other side) we go to equaton (1) and replace x for its value as function of L
y = ( L - x )/2 y = ( L - L/2)/2 y = ((2L-L)/2))/2 y = ( 2L-L )/ 4
y = L/4
And maximum area would be Amax. = x × y = L/2 × L/4 Amax. = L^{2}/8 square feet
