Solve each of the following linear programming problem by ... - Free Printable

Let’s solve this step by step.

We are asked to maximize the function:
> Z = 18x + 10y

Subject to these constraints:
1. 4x + y ≤ 20
2. 2x + 3y ≤ 30
3. x ≥ 0, y ≥ 0 (since it says x, y > 0 — but in linear programming, we usually include 0 as a boundary point)

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Step 1: Find the corner points of the feasible region

The feasible region is the area that satisfies all inequalities. The maximum (or minimum) of a linear function like Z always occurs at one of the corner points (vertices) of this region.

From the graph and calculations shown, we can see the corner points are:

- (0, 0) → origin
- (5, 0) → where 4x + y = 20 meets the x-axis (set y=0 → 4x=20 → x=5)
- (0, 10) → where 2x + 3y = 30 meets the y-axis (set x=0 → 3y=30 → y=10)
- Point P(3, 8) → intersection of the two lines: 4x + y = 20 and 2x + 3y = 30

Let’s verify point P(3,8):

Plug into first equation:
4*(3) + 8 = 12 + 8 = 20 ✔

Plug into second equation:
2*(3) + 3*(8) = 6 + 24 = 30 ✔

So yes, (3,8) is correct.

But wait — what about (0,0)? Let’s check if it’s included. Since x,y ≥ 0, and both constraints are satisfied at (0,0), it should be considered too.

However, looking at the shaded region in the sketch, it seems they only considered the bounded polygon formed by (5,0), (3,8), and (0,10). But actually, (0,0) is also part of the feasible region.

BUT — let’s think: since we’re maximizing Z = 18x + 10y, and both coefficients are positive, the maximum will NOT occur at (0,0) because increasing x or y increases Z. So even if we include (0,0), its Z value will be 0, which is definitely not the max.

Still, for completeness, let’s list ALL corner points of the feasible region:

Actually, let’s plot mentally:

Constraint 1: 4x + y ≤ 20 → line from (5,0) to (0,20)
Constraint 2: 2x + 3y ≤ 30 → line from (15,0) to (0,10)

But since both must hold, the feasible region is bounded by:

- From (0,0) up along y-axis to (0,10) [because beyond y=10, constraint 2 fails]
- Then along 2x+3y=30 to point P(3,8)
- Then along 4x+y=20 down to (5,0)
- Then back to (0,0) along x-axis? Wait — no! At (5,0), check constraint 2: 2*5 + 3*0 = 10 ≤ 30 → OK. At (0,10): 4*0 + 10 = 10 ≤ 20 → OK.

Wait — actually, the feasible region is a quadrilateral with vertices:

→ (0,0)
→ (5,0)
→ (3,8)
→ (0,10)

Yes! Because:

- Between (0,0) and (5,0): along x-axis, under both constraints.
- Between (5,0) and (3,8): along 4x+y=20
- Between (3,8) and (0,10): along 2x+3y=30
- Between (0,10) and (0,0): along y-axis

So four corner points: (0,0), (5,0), (3,8), (0,10)

Now compute Z = 18x + 10y at each:

1. At (0,0): Z = 18*0 + 10*0 = 0
2. At (5,0): Z = 18*5 + 10*0 = 90
3. At (3,8): Z = 18*3 + 10*8 = 54 + 80 = 134
4. At (0,10): Z = 18*0 + 10*10 = 100

So the values are: 0, 90, 134, 100

Clearly, the maximum is 134 at point (3,8)

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Wait — the original handwritten solution says “Hence, z is maximum at points (3,8) & have maximum value = 134” — which matches our calculation.

But note: they didn’t include (0,0) in their table — probably because it’s obvious it won’t give max. That’s fine.

Also, double-check arithmetic at (3,8):

18 * 3 = 54
10 * 8 = 80
54 + 80 = 134 → Correct.

At (0,10): 10*10=100 → Correct.

At (5,0): 18*5=90 → Correct.

So everything checks out.

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Final Answer:
The maximum value of Z is 134, achieved at the point (3, 8).