Consider the following problem: a farmer with 850 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. what is the largest possible total area of the four pens? (a) draw several diagrams illustrating the situation, some with shallow, wide pens and some with deep, narrow pens. find the total areas of these configurations. does it appear that there is a maximum area? if so, estimate it.

The maximum area is 18062.5 ft².

Step-by-step explanation:

Consider the provided information.

A farmer with 850 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle.

The required figure is shown below.

The total length of wire is 850

The perimeter is the sum of all side which is the same as the total length of the wire, therefore

Total fence = Length of 3 parallel wall + Sides of rectangle

The area of rectangle is A=x×y

Substitute the value of y.

Now differentiate and substitute A'=0 to find the maximum.

Now substitute the value of x in

We need to find the total areas of theses configurations.

Hence, the maximum area is 18062.5 ft².

step-by-step explanation:

why? because i noticed a pattern. 60 120. those numbers without the zero, make the 6 times tables leaving your answer to be b since 6 times 4 is 24 and adding a zero making it 240.