Two part Question (a) Verify that a rectangle with one side x and perimeter k has an area A = x (k-2x/2) (b) In the...

To verify the formula, A=x(k-2x/2), where  A=area, k = perimeter and x is one side of the rectangle.

Verification:

If x=2 and the other side of  the rectangle is 3, the area =side* other side =2*3 = 6....................................(1)

k=(2+3)2=10

k =  10 . Then the area by the given formula, A = 2(10-2*2/2) = 2*8=16, which does not tally with the result side*other side  result at (1).

But , when x=2, and k=10,  the other side is  (k-2*2)/2 = (10-2*2)/ = 6/2=3. And area 2*3=6. So the given formula does not hold good. It needs a correction.

The other side should be (k-2x)/2 and not (k-2x/2). The formula for area should be A = x(k-2x)/2. Now substitute x=2 and k=10 in A =  x(k-2x)/2  and you get A = 2(10-2*2)/2 =2*(10-4)/2 = 2*3 = 6.

b)

Shape of the plot is rectangular. Therefore area of the plot is given by:

A(x) = x*(k-2x)/2 , k is the perimeter fixed and x is variable.

A(x) is maximum for  for that value of x for which A'(x) = 0 and A''(x) = negative.

Therefore, A'(a)=0 gives: [x*(k-2x)/2]'= 0 or{(kx-2x^2)/2}' = 0 or

(k-4x)/2=0 or x= k/4.

A"(k/4) = {( k-4x)/2}' at x=k/4

=-4/2 = -2, which is negative.

Therefore, x= k/4. Here k is given 1.6m =1600 m

Therefore, x= 1600/4=400m and A = 400^2 sq m=160000 sq m