Use calculus to find the largest possible area for a rectangular field that can be enclosed with a fence that is 500 meters long.

The maximum area that can be enclosed is 15625 feet²

A rectangle is a quadrilateral in which opposite sides are equal and parallel to each other.

Let x represent the length and y represent the width. Hence:

perimeter = 2(length + width)

500 = 2(x + y)

250 = x + y

y = 250 - x

The area of a rectangle (A) = length * width

A = x * (250 - x)

A = 250x - x²

The maximum area is at dA/dx = 0, hence:

dA/dx = 250 - 2x

250 - 2x = 0

2x = 250

x = 125 feet

y = 250 - x = 250 - 125 = 125 feet

The maximum area = x * y = 125 * 125 = 15625 feet²

The maximum area that can be enclosed is 15625 feet²

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step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

the car was traveling 40 mph before it started skidding. ( 2 points ) 17. was the driver were her brakes working at 75% efficiency or better? ( 2 points ) yes 29.

step-by-step explanation: