Ineed it’s due

a farmer is enclosing a rectangular area for a pigpen. he wants the length of the pen to be 15 ft longer then the width. the farmer can only afford no more than 150ft of fencing. you draw the length of the pigpen on paper and “use” the picture i have made

Let
x-------> the width of the rectangular area
y------> the length of the rectangular area

we know that
y=x+15------> equation 1
perimeter of a rectangle=2*[x+y]
2x+2y <= 150-------> equation 2

substitute 1 in 2
2x+2*[x+15] <=150--------> 2x+2x+30 <=150----> 4x <=150-30
4x <= 120---------> x <= 30
the width of the rectangular area is at most 30 ft
y=x+15
for x=30
y=30+15------> y=45
 the length of the rectangular area is at most 45 ft

see the attached figure
the solution is the shaded area

Iwould say d but i am not 100% positive