The city park department is planning to enclose a play aea. one side of the area will be against an existing building, so no fence is needed there. find the dimensions ofthe maximum rectangular area that can be enclosed with 800m of fence

length = 200 m

width = 400 m

Step-by-step explanation:

Let the length of the plaing area is L and the width of the playing area is W.

Length of fencing around three sides = 2 L + W = 800

W = 800 - 2L (1)

Let A is the area of playing area

A = L x W

A = L (800 - 2L)

A = 800 L - 2L²

Differentiate with respect to L.

dA/dL = 800 - 4 L

It is equal to zero for maxima and minima

800 - 4 L = 0

L = 200 m

W = 800 - 2 x 200 = 400 m

So, the area is maximum if the length is 200 m and the width is 400 m.

Dimensions : Length = 400 , Width = 200

Step-by-step explanation:

Let length & width of rectangle park be = L , W . Also, let one side of length be supported by building wall, so not needing fencing.

So, the perimeter of rectangle park, including 2 width & 1 length :

2W + L = 800

L = 800 - 2W

Rectangle Area = Length x Width

A = L x W

A = (800 - 2W) W

A = 800W - 2W^2

To maximise area, it will have to be differentiated w.r.t dimension width

dA / dW = 800 - 4W

dA / dW = 800 - 4W  = 0

800 = 4W

W = 800 / 4

Width [W]  = 200

Length [L] = 800 - 2W

= 800 - 2(200)

= 800 - 400

Length [L] = 400

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.