if a,b,c are positive reals and a+b+c= 50 and 3a+b-c=70.If ex=5a+4b+2c, find the range of values of x

Question

i need a solution

Answer 1

We can find the line of intersection of the 2 planes a+b+c = 50; 3a+b-c = 70 by calculating the cross-product.

 

i  j   k

1 1  1

3 1 -1

 

We get -2i + 4j - 2k, which we can reduce to i - 2j + k

 

Then, find a particular value.   Let a = 0; Then, b+c = 50; b - c = 70; this gives a value b = 60; c = -10; while this is invalid, we won't let this stop us.

Applying i - 2j + k, the direction of the line is 1, -2, 1; Our solution is (a, 60 - 2a, a - 10)  

Note how that works with both a+b+c = 50 and 3a + b - c = 70

As a, b, and c are positive reals, for c >= 0, a - 10 >= 0, a >= 10.

Then, b = 60 - 2a implies for b >= 0, a <= 30

Thus, 10 <= a <= 30

 

Now, the objective function is 5a + 4b + 2c = 5a + 4(60 - 2a) + 2(a - 10) = 5a + 240 - 8a + 2a - 20 = 220 - a

Clearly, this decreases with increasing a.

Thus, the maximum is at a = 10, and the minimum is at a = 30

Maximum = 220 - 10 = 210

Minimum = 220 - 30 = 190