Answer:
area = 122 sq unit
Step-by-step explanation:
Given constraints:
5X + 5Y < 80
2X + 6Y < 72
3X + 2Y < 42
X , Y > 0
To find the co-ordinates we can find
[tex]\dfrac{x}{16} +\dfrac{y}{16} <1\\\dfrac{x}{36} +\dfrac{y}{12} <1\\\dfrac{x}{14} +\dfrac{y}{21} <1[/tex]
the co-ordinates are shown in the diagram
by solving equation
5X + 5Y < 80
3X + 2Y < 42
we will get the intersection point x = 10 and Y = 5
shaded region in the graph shows the required region
required area can be found out by
[tex]area = \int_{0}^{10}(16-x)dx+\int_{10}^{14}(21-1.5x)dx[/tex]
[tex]area=\left ( 16x-\frac{x^2 }{2}\right)^{10}_0 +\left ( 21x-\frac{1.5x^2 }{2}\right)^{14}_{10}[/tex]
area = 122 sq unit