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Application Of Integrals

Question
CBSEENMA12035687

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

Solution
straight x minus 2 straight y less-than or slanted equal to 2
3 straight x plus 2 straight y less-than or slanted equal to 12
minus 3 straight x plus 2 straight y less-than or slanted equal to 3
straight x greater-than or slanted equal to 0 comma space straight y greater-than or slanted equal to 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.