Respuesta :
Let the width of the corral be x and the length of the whole (undivided) corral be y. We want to maximize the area. A = xy.
He has 500 feet of fencing, 3x of which is taken up by the shorter ends and the dividing wall of the corral, and 2y of which is taken up by the longer sides.
Then 3x+2y=500 ft. Supposing that we choose to solve for x first: 2y=500-3x, or
500-3x
y = ------------
2
500-3x
Then the formula for the area becomes A = x[-----------]
2
2A = 500x -3x^2. We must differentiate both sides with respect to x:
2(dA/dx) = 500 - 6x and seet this derivative equal to zero to find the critical value(s):
0 = 500 - 6x becomes 6x = 500, or x = 250/3, or 83 1/3 feet. Earlier, we saw that
500-3x
y = ------------
2
so, to calculate y, we substitute x=83 1/3 feet into that formula:
500-3(250/3)
y = -------------------- ft = 250/2 ft = 125 ft.
2
The area is then A = xy, or (83 1/3 ft)(125 ft) = 10,416 2/3 square feet.
He has 500 feet of fencing, 3x of which is taken up by the shorter ends and the dividing wall of the corral, and 2y of which is taken up by the longer sides.
Then 3x+2y=500 ft. Supposing that we choose to solve for x first: 2y=500-3x, or
500-3x
y = ------------
2
500-3x
Then the formula for the area becomes A = x[-----------]
2
2A = 500x -3x^2. We must differentiate both sides with respect to x:
2(dA/dx) = 500 - 6x and seet this derivative equal to zero to find the critical value(s):
0 = 500 - 6x becomes 6x = 500, or x = 250/3, or 83 1/3 feet. Earlier, we saw that
500-3x
y = ------------
2
so, to calculate y, we substitute x=83 1/3 feet into that formula:
500-3(250/3)
y = -------------------- ft = 250/2 ft = 125 ft.
2
The area is then A = xy, or (83 1/3 ft)(125 ft) = 10,416 2/3 square feet.