A man has 340 yds.of fencing to enclose 2 separate fields,1 a rectangle 2x as long as its wide and 1 a square?

The square field must contain at least 100sq yds and the rectangular field must contain at least 800sq yds. if x is the width of the rectangular firld, what are the max and min possible values of x? What is the greatest number of square yards that can be enclosed in the two fields? Justify your answer and use calculus.|||To find the minimum width of the rectangular field, we note that the minimum area of the rectangular field is 800.


A = W*L


800 = x * 2x


800 = 2x^2


400 = x^2


x = sqrt(400)


x = 20





So the rectangular field has the width of 20 Minimum.





To find the maximum width of the rectangular field, first we find out how many yards of fencing we have to use for the square field. The area of the sqare field is minimum 100, therefore:


A = S^2


S = sqrt(A)


S = aqrt(100)


S = 10





So each side of the square field is 10 yards. The perimeter of the square field is:


P = 4S


P = 4*10


P = 40





So we use 40 yards of fencing to enclose the square field. That leaves us with 340-40 = 300 yards of fencing. The perimeter of the square field is:


P = 2*(L+W)


300 = 2(2x + x)


300 = 2(3x)


300 = 6x


x = 50





So the rectangular field has the width of 50 Maximum.





Now, we have 340 yards of fencing to enclose both fields, so:


340 = Perimeter of square + Perimeter of rectangle


340 = (4*S) + 2*(L+W)


340 = 4S + 2*(2x+x)


340 = 4S + 2*(3x)


340 = 4S + 6x





Let%26#039;s solve for S


4S = 340 - 6x


S = (340 - 6x) / 4


S = 340/4 - 6/4x


S = 85 - 3/2x





The total area of the two fields is:


A = Area of Square + Area of Rectangle


A = S^2 + (x*(2x))


A = (85 - 3/2x)^2 + 2X^2


A = 85^2 + (3/2x)^2 - 2*85*3/2x + 2x^2


A = 7225 + 9/4x^2 - 255x + 2x^2


A = 17/4x^2 - 255x + 7225





If you graph this formula, you see that this is a U-shaped graph. Finding the minimum area is easy, just get the first derivative and solve for x (which will give you x = 30), but since we are looking for the maximum area, and the area increases as you move away from 30 to either direction, and we know that the min and max of x are 20 and 50, the best thing to do is to plug in both values for x and see which is higher


f(x) = 17/4x^2 - 255x + 7225





f(20) = 17/4(20)^2 - 255(20) + 7225


f(20) = 1700 - 5100 + 7225


f(20) = 3825





f(50) = 17/4(50)^2 - 255(50) + 7225


f(50) = 10625 - 12750 + 7225


f(50) = 5100





So a rectangular field of 50x100 and a square field of 10x10 gives us the greatest square yardage|||Getting you started:


x = length of square area


y = shorter length of rectangular area





length equation:


(4x) + (2y + 4y) = 340





area equations:


x^2 %26lt;=100 x%26lt;=10


(2y^2) %26lt;= 800 y^2 %26lt;=400 y%26lt;=20





One answer:


Square yard (x) is 10x10 (40 linear feet)


Rectangular yard (y) is 20 x40 (120 linear feet)


Total linear feet = 160%26#039;