Farmer ed has 700 meters of fencing, and wants to enclose a rectangular plot that borders on a river. if farmer ed does not fence the side along the river, find the length and width of the plot that will maximize the area. what is the largest area that can be enclosed?

Length = 350 meters

Width = 125 metres

Maximum area = 43750 meter²

Step-by-step explanation:

Farmer Ed has total fencing = 700 meters

Let the length of rectangular plot is = x meters

Let the width of the plot is = y meters

Since fencing is to be done on three sides so x + 2y = 600 -------(1)

Now the area of plot A = xy --------(2)

Now we substitute the value of x in the equation 2 from equation 1.

A = y(600 - 2y)

A = 600y - 2y²

For the maximum area we find the derivative of the plot which will be equal to the zero.

= 600 - 4y

600 - 4y = 0

4y = 600

y = 125 meters

By putting y = 125 in equation 1

x + 2(125) = 600

x + 250 = 600

x = 600 - 250

x = 350 meters

So the length will be 350 meters and width will be 125 meters

And the largest area enclosed will be = 350×125 = 43750 meter²

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.