minimal value of 3x + 5y?

Question

what is the minimal value of 3x + 5y subject to the constraints:
3x +2y ≥ 36
3x + 5y ≥ 45
x ≥ 0
y ≥ 0
and what are the points in this region at which this minimum is obtained.

Answer 1

3x +2y ≥ 36 ... A
3x + 5y ≥ 45 ... B
x ≥ 0 ... C
y ≥ 0 ... D

From B, we know that the minimal value is at least 45. Assume it is 45. Then 3x=45-5y.

Plugging into A, we get 45-3y≥36. Thus, 3≥y.

Combining with equation D, 3≥y≥0.

Now we need only to check that C holds. From 3≥y≥0 and 3x=45-5y, we get 15≥x≥10, so C is always true in this range.

So, the region is the line 3x+5y=45, with range 3≥y≥0, and domain 15≥x≥10.

Answer 2

There are two ways to go about this problem: the hard method, which involves carefully considering the border of the region, and evaluating it at all critical points to find where the minimum of the function might be, or the easy method, which consists of the following:

3x + 5y ≥ 45, therefore the minimum of 3x+5y is 45.

That being extremely easy, now we must find out: which of the points on the line where 3x+5y=45 are actually IN the region (i.e. satisfy the other three constraints)? We must solve for the points of intersection with the other lines. The two axes are trivial to solve, and for the third line:

3x+5y=45
3x+2y=36

Subtract the second equation from the first:

3y=9
y=3

3x+5(3)=45
3x=30
x=10

And simple testing reveals that if 3x+5y=45, then 3x+2y>36 for x>10. So the points we are considering are those on 3x+5y=45 where x>10. There is thus no need to consider the intersection with the x-axis. For the y-axis:

3x=45 → x=15

Thus the points on 3x+5y=45 which satisfy the other constraints are those on the line segment from (10, 3) to (0, 15), or formally:

{(x, y): 3x+5y=45 ∧ 10≤x≤15}

Answer 3

The answer is right in front of you:

Minimal value of 3x + 5y is 45
when x=10 and y = 3