Answer to Question #135681 in Operations Research for Cameron Foster

Question

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Find the Min and Max of Z = 5x + 3y
subject to:
4x + 5y ≤ 200
3x + 6y ≤ 180
x ≥ 0; y ≥ 0

Answer 1

here,given function is z=5x+3y.

we have to find minimum and maximum in below region:

1)4x+5y≤200

2)3x+6y≤180

3)x≥0

4)y≥0

the region cover by above inequality is by point

(0,0),(0,30),(33.333,13.333),(50,0).

to find minimum and maximum we put the above point value in z=5x+3y

(i)for(0,0) we get z=0.

(ii)for (0,30) we get z=90.

(iii)for (33.333,13.333) we get z=206.664

(iv)for (50,0) we get z=250.

from above 4 point value we can say thay z=5x+3y attains it's maximum value 250 at(50,0) and minimum value 0 at (0,0) in given region.

To find the intersect point of (i)4x+5y≤200(ii)3x+6y≤180(iii)x≥0(iv)y≥0

We take equality instead of inequality to find intersection point of 4x+5y=200 and 3x+6y=180

Let set 4x+5y=200 as equation 1.

And 3x+6y=180 as equation 2.

Take 3 common from the equation 2 we get x+2y=60.and set x+2y=60 as equation 3.

By equation 3 we can get value of x=60-2y.

Put the value of x in equation 1.

We get 4(60-2y)+5y=200

240-8y+5y=200

-3y=-40

y=13.333

Put the value of y in equation 1 we get

4x+5*(13.333)=200

x=33.333

We get(33.333,13.333) as a intersection point

For z=5x+3y we get 206.64(rounded) at(33.333,13.333)

It is less than 250 and 250 is attained at (50,0)

So z=5x+3y attains maximum at (50,0) and also z= 5x+3y attains minimum 0 at (0,0).