Having a hard time solving this. Help pls.

[vanna]

Max P = 4x+6y

Subject to:

3x+8y[tex]\le[/tex]48
3x+2y [tex]\le[/tex]24
x,y[tex]\ge[/tex]0


Replies pls. Thanks.

[Guest]

3x+8y ≤ 48
3x+2y ≤ 24
Subtract gives 6y <= 24
y <= 4

3x + 2y <= 24
3x + 8 <= 24
3x <= 16
x <= 5.33

The point (x,y) = (5.33, 4) is the point of intersection of the lines 3x + 8y = 48 and 3x+25 = 24 the inequalities include these lines and below them for x,y >= 0
for Max P then line P= 4x+6y passes through this point, So, P = 4x5.33 + 6x4 = 45.32

Max P = 45.32

[leesajohnson]

3x+8y ≤ 48
3x+2y ≤ 24
Subtract gives 6y <= 24
y <= 4

3x + 2y <= 24
3x + 8 <= 24
3x <= 16
x <= 5.33
Now put the value of x and y into the equation P= 4x+ 6y

P= 4 x 5.33 + 6 x 4
P= 21.32 + 24
P= 45.32

[Guest]

Both of the posts above give the right answer but are nonsensical.

The inequalities [tex]3x+8y\leq 48[/tex] and [tex]3x+2y\leq 24[/tex] cannot be subtracted to imply [tex]6y\leq 24[/tex]. For example take [tex]x=0[/tex] and [tex]y=5[/tex], these satisfy the original inequalities but not the "subtracted" inequality.

The correct way to solve this is by something like the simplex algorithm, which will say that the objective function is maximised at the intersection of the boundaries of two of the inequalities (or be unfeasible or unbounded). In your case it will be when [tex]3x+8y = 48[/tex] and [tex]3x+2y = 24[/tex] which implies [tex]x=5\tfrac{1}{3}, y=4[/tex] which implies the maximum value of the objective function is [tex]45\tfrac{1}{3}[/tex].

Hope this helped,

R. Baber.

[Guest]

Another way of solving it is the following:

Given a fixed [tex]y\geq 0[/tex], what do the constraints on [tex]x[/tex] look like.
[tex]3x+8y\leq 48[/tex] becomes [tex]x\leq 16-\tfrac{8}{3}y[/tex]
[tex]3x+2y\leq 24[/tex] becomes [tex]x\leq 8-\tfrac{2}{3}y[/tex]
For there to be any feasible solutions we need [tex]0\leq x[/tex] which implies
[tex]0\leq 16-\tfrac{8}{3}y[/tex] which means [tex]y\leq 6[/tex]
and
[tex]0\leq 8-\tfrac{2}{3}y[/tex] which means [tex]y\leq 12[/tex] (which is a weaker condition than [tex]y\leq 6[/tex] so we can ignore it).

So now we have [tex]0\leq y \leq 6[/tex] and [tex]0\leq x \leq \min\{16-\tfrac{8}{3}y,8-\tfrac{2}{3}y\}[/tex].
For a fixed [tex]y[/tex] to maximize the objective function we should take [tex]x[/tex] as large as possible, so [tex]x =\min\{16-\tfrac{8}{3}y,8-\tfrac{2}{3}y\}[/tex].

There are now two cases to consider (depending on which of [tex]16-\tfrac{8}{3}y[/tex] and [tex]8-\tfrac{2}{3}y[/tex] is smaller).

Case 1:[tex]\quad 16-\tfrac{8}{3}y\leq 8-\tfrac{2}{3}y[/tex]
This implies [tex]x=16-\tfrac{8}{3}y[/tex], making the objective function
[tex]4(16-\tfrac{8}{3}y)+6y = 64-\tfrac{14}{3}y[/tex]
So we should choose [tex]y[/tex] as small as possible.
Also [tex]16-\tfrac{8}{3}y\leq 8-\tfrac{2}{3}y[/tex] rearranges to [tex]4\leq y[/tex] making the constraints on [tex]y[/tex]
[tex]4\leq y \leq 6[/tex].
So the optimal solution in this case is [tex]y=4[/tex], [tex]x=5\tfrac{1}{3}[/tex].

Case 2:[tex]\quad 16-\tfrac{8}{3}y\geq 8-\tfrac{2}{3}y[/tex]
This implies [tex]x=8-\tfrac{2}{3}y[/tex], making the objective function
[tex]4(8-\tfrac{2}{3}y)+6y = 32+\tfrac{10}{3}y[/tex]
So we should choose [tex]y[/tex] as large as possible.
Also [tex]16-\tfrac{8}{3}y\geq 8-\tfrac{2}{3}y[/tex] rearranges to [tex]4\geq y[/tex] making the constraints on [tex]y[/tex]
[tex]0\leq y \leq 4[/tex].
So the optimal solution in this case is [tex]y=4[/tex], [tex]x=5\tfrac{1}{3}[/tex].

So the optimal solution in both cases is [tex]y=4[/tex], [tex]x=5\tfrac{1}{3}[/tex] giving [tex]P=45\tfrac{1}{3}[/tex].

Hope this helped,

R. Baber.