Amaths

1) Let the length of a square be x, A be the total areas enclosed.
The length of the other square = (8 - 4x)/4 = 2 - x
A = x² + (2-x)²
= 2x² - 4x + 4
dA/dx = 4x - 4
d²A/dx² = 4
For A to be minimum, dA/dx = 0 AND d²A/dx² > 0 for that value of x.
dA/dx = 4x - 4 = 0
x = 1
d²A/dx²(at x = 1) = 4 > 0
So A is minimum when x = 1.
The length of the other square = 2 - x = 1
So the lengths of the two squares are both 1cm for the total area enclosed to be minimum.
b) Area of PQRS = Area of ABCD - Area of APS - Area of BQP - Area of CRQ - Area of DSR
= 8*6 - 0.5*x*(6-x) - 0.5*x*(8-x) - 0.5*x*(6-x) - 0.5*x*(8-x)
= 48 - x(6-x) - x(8-x)
= 2x² - 14x + 48
Let R be the area of PQRS
R = 2x² - 14x + 48
For R to be minimum,
dR/dx = 0
4x - 14 = 0
x = 7/2
d²R/dx² = 4 > 0
So the area of PQRS is minimum when x = 7/2.

1)
Suppose the wire is cut into portions with lengths x cm and (8-x) cm with 0 < x < 8.
Then the squares will have their side as (x/4) cm and (2-x/4) cm
So the total area is given by:
(x/4)^2 + (2-x/4)^2
= x^2/16 + 4 - x + x^2/16
= x^2/8 - x + 4
= (1/8)(x^2 - 8x) + 4
= (1/8)(x^2 - 8x + 16) + 4 - (1/8)(16)
= 2 + (1/8)(x - 4)^2

So the area will be minimum when x = 4 cm and the minimum area is 2 cm^2.

2)
With 0 < x < 6, 4 smaller right-angled triangles are cut from the original rectangle are cut out. 2 with dimensions (8-x) cm by x cm and the other 2 with dimensions (6-x) cm by x cm.
So the area of the parallelogram is given by:
48 - (8-x)x - (6-x)x
= 48 - 8x + x^2 - 6x + x^2
= 2x^2 - 14x + 48
= 2(x^2 - 7x) + 48
= 2[x^2 - 7x + (7/2)^2] + 48 - 2(7/2)^2
= 2(x - 7/2)^2 + 48 - 49/2
= 47/2 + 2(x - 7/2)^2
So the area will be minimum when x = 3.5 cm and the minimum area is 23.5 cm^2.