The city park department is planning to enclose a play area with fencing. One side of the area will be against an existing building, so no fence is needed there. Find the dimensions of the maximum rectangular area that can be enclosed with 800 meters of fence. Include the units.

The maximum rectangular area will have the length 400 meters and width 200 meters with one side of the length against an existing building.

Step-by-step explanation:

From the given information;

The perimeter of a rectangle = 2 (L+B)

here;

L = the length of the side of the fence

B = the width of the fence

So; The perimeter of a rectangle = 2L+2B

we are also being told that;

One side of the area will be against an existing building

i.e

The perimeter of a rectangle is now =  L + 2B = 800 meters

L = 800 - 2B

Similarly; Area of a rectangle = L × B

Area of a rectangle = ( 800 - 2B) × B

Area of a rectangle =  800B - 2B²

assuming A(B) to represent the Area;

Then the maximum area A'(B) = 0 ;

Thus,

A'(B) = 800 - 4B = 0

-4B = - 800

4B = 800

B = 200

Therefore; the maximum area have a width = 200 meters and a length 800 - 2(200) =  800 - 400 = 400 meters

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

step-by-step explanation:

this is not possible to solve with the info given.   you need at least an one angle somewhere.