A rectangular field is built next to a house and is enclosed by 400 m of fencing for the 3 remaining sides. Determine the dimensions of the field to maximize the area to nearest tenth of a metre.

The dimensions of the field to maximise the area are 100 metres (width) and  200 metres (height).

Step-by-step explanation:

The formulas for the area (), measured in square metres, and the perimeter (), measured in metres, of the rectangle are, respectively:

(1)

(2)

Where:

- Width, measured in metres.

- Height, measured in metres.

Note: We assume that height of the rectangle is parallel to the wall of the house.

By (2):

In (1):

(3)

Then, we obtain its first and second derivatives by Differential Calculus:

(4)

(5)

By equalising (4) to zero, we find the following critical value for :

And besides the Second Derivative Test, this solution is associated to an absolute maximum. Given that , then the maximum area enclosed by fencing is:

And the height of the triangle is:

The dimensions of the field to maximise the area are 100 metres (width) and  200 metres (height).

so a parallelogram is just like a square base times height. multiply 40 and 39 to get 1560 inches.

a triangle is base times height divided by two. so multiply 7.9 and 5 and then divide it by 2 to get 19.75inches.

the kite can be split into two separate triangles and added together to get the total area. 30 x 4 /2 which equals 60. make sure to double it for the two triangles making 120.

i'm sorry mate, but i am not seeing any figures. i am unable to solve this for you.