The vertices of a feasible region are (1, 2) (7, 3) and (5, 1). What is the minimum value of the function P = x - 2y?

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24-08-2022
Mathematics

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The vertices of a feasible region are (1, 2) (7, 3) and (5, 1). What is the minimum value of the function P = x - 2y?

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priyankapandeyVT priyankapandeyVT · Answer 1 · 2022-08-26T14:51:04+02:00

The required minimum value of P = x - 2y is -3.

Given that,
The vertices of a feasible region are (1, 2) (7, 3), and (5, 1). What is the minimum value of the function P = x - 2y is to be determined.

What are functions?

Functions are the relationship between sets of values. e g y=f(x), for every value of x there is its exists in a set of y. x is the independent variable while Y is the dependent variable.

Here,
For (1, 2)
P = x - 2y
P = 1 - 2*2
P = -3

For, (7, 3)
P = x - 2y
P = 7 - 2 *3
P = 1

For (5, 1)
P = x - 2y
P = 5 - 2
P = 3

Thus, the required minimum value of P = x - 2y is -3.

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