Afarmer has 300 ft of fencing with which to enclose a rectangular pen next to a barn. the barn itself will be used as one of the sides of the enclosed area. what is the maximum area that can be enclosed by the fencing? enter your answer in the box. ft²

Let
y------> the length of the rectangle
x-----> the width of the rectangle

we know that
perimeter of the rectangle=2*[x+y]

Since the barn is used as one of the sides (let's say y) we can subtract y
we don't need fencing for this side
That makes the perimeter 2x + y
Since we have 300 feet of fencing
we set these equal:
 2x + y = 300-------> y=(300-2x)------> equation 1

area of the rectangle=x*y
substitute the equation 1 in the area formula
Area=[(300-2x)]*x-----> Area=(-2x²)+300x

This is a quadratic equation
Since the leading coefficient is negative (-2) we know it opens downward 
We are looking for the x-coordinate of the highest point called the vertex

using a graph tool
see the attached figure

the vertex is the point (75,11250)
that means for x=75 (width of the rectangle)

the area is 11250 ft²

the answer is
the maximum area that can be enclosed by the fencing is 11250 ft²

You need to divide 210 square inches by 35 inches and that’s going to give you 6 inches and that’s the length of the hypothenuse of the right triangle i hope that this you : )