# Two part Question - Calculus (a) Verify that a rectangle with one side x and perimeter k has an area A = x (k-2x/2) (b) In the early days of settlement , settlers were permitted...

Two part Question - Calculus

(a) Verify that a rectangle with one side *x *and perimeter *k *has an area

A = x (k-2x/2)

(b) In the early days of settlement , settlers were permitted to select a homestead block, a rectangular area of crown land that had a perimeter of approximately 1.6 km. Use calculus to find the shape of the block that gave the greatest area to the selection and give the maximum size of this selection in square metres.

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A rectangle has four sides with measures of opposite sides being equal. Thus the perimeter of a rectangle (k) is given by:

k = 2*(Measure of one side + Measure of other side)

If measure of one side is x, then this equation becomes:

k = 2*(x + Measure of other side)

Therefore:

Measure of other side = (k/2 - x) = (k - 2x)/2

Area of rectangle = A = (Measure of one side)*(Measure of other side)

Substituting value of the two sides in the above equation we get:

A = x*[(k - 2x)/2]

We can simplify this further as:

A = kx/2 - x^2

When value of k is 1.6 km this equation becomes:

A = (1.6x)/2 - x^2 = 0.8x - x^2

To find the value of x when the area (A) is maximum we find the differential of the right hand side of above equation. A will be maximum when this is zero.

Therefore: 0.8 - 2x = 0

Or: x = 0.4

Thus one side of the rectangle will measure 0.4 km.

And the other side will measure (k/2 - x) = 1.6/2 - 0.4 = 0.4

Area of this rectangle

A = (Measure of one side)*(Measure of other side) = 0.4*0.4 = 0.16 km^2

Or = 0.16*100,000 m^2 = 160.000 m^2

Answer:

Block that gives maximum area will be a square with each side equal to 0.4 km.

The total area of this block is 160.000 m^2

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