A retired person wants to invest an amount of Rs. 50, 000. His broker recommends investing in two type of bonds ‘A’ and ‘B’ yielding 10% and 9% return respectively on the invested amount. He decides to invest at least Rs. 20,000 in bond ‘A’ and at least Rs. 10,000 in bond ‘B’. He also wants to invest at least as much in bond ‘A’ as in bond ‘B’. Solve this linear programming problem graphically to maximise his returns.

Minimum and maximum z = 5x + 2y subject to the following constraints:
x – 2y ≤ 2
3x + 2y ≤ 12
−3x + 2y ≤ 3
x ≥ 0, y ≥ 0

Converting the inequations into equations, we obtain the lines
x – 2y = 2…..(i)
3x + 2y = 12……(ii)
−3x + 2y = 3……(iii)
x = 0, y = 0

From the graph, we get the corner points as
A(0, 5), B(3.5, 0.75), C(2, 0), D(1.5, 3.75), O(0, 0)
The values of the objective function are: 

Point (x,y)	Values of the objective function Z= 5x+2y
A(0, 5)	5 × 0 + 2 × 5 = 10
B(3.5, 0.75)	5 x 3.5 +2 x 0.75 =19 (Maximum)
C(2, 0)	5 x 1.5 +2 x 3.75 =15
O(0,0)	5 x 0 + 2 x 0 = 0 (Minimum)

The maximum value of Z is 19 and its minimum value is 0.

Using integration, find the area of the region bounded by the triangle whose vertices are (-1, 2), (1, 5) and (3, 4).

Using integration, find the area bounded by the curve x² = 4y and the line x = 4y – 2.