Formulate four inequalities to represent the given conditions and show the feasible region on a graph.

• "at least 10 dining chairs":

• "at least 20 garden chairs":

• "not more than 80 chairs altogether":

• "number of garden chairs must not be more than three times the number of dining chairs":

The four inequalities are:

Label the x-axis as "Number of Dining Chairs" and the y-axis as "Number of Garden Chairs".

2cm represents 10 chairs on both axes. This means that for every 2cm interval, mark 10, 20, 30, ..., up to 80.

• For $x \ge 10$, draw the vertical line $x = 10$.
• For $y \ge 20$, draw the horizontal line $y = 20$.
• For $x + y \le 80$, draw the line $x + y = 80$. This line passes through $(80, 0)$ and $(0, 80)$.
• For $y \le 3x$, draw the line $y = 3x$. This line passes through $(0, 0)$, $(10, 30)$, $(20, 60)$, etc.

• $x \ge 10$: The region to the right of $x = 10$.
• $y \ge 20$: The region above $y = 20$.
• $x + y \le 80$: The region below the line $x + y = 80$.
• $y \le 3x$: The region below the line $y = 3x$.

This means shading the areas that do not satisfy the inequalities.

• Shade the region to the left of $x = 10$.
• Shade the region below $y = 20$.
• Shade the region above $x + y = 80$.
• Shade the region above $y = 3x$. The unshaded region will be the feasible region, which is a polygon with vertices at the intersections of these lines. The vertices of the feasible region are:
• Intersection of $x=10$ and $y=20$: $(10, 20)$
• Intersection of $x=10$ and $y=3x$: $(10, 30)$
• Intersection of $y=3x$ and $x+y=80$: $x+3x=80 \implies 4x=80 \implies x=20$. Then $y=3(20)=60$. So, $(20, 60)$
• Intersection of $y=20$ and $x+y=80$: $x+20=80 \implies x=60$. So, $(60, 20)$

Profit on a dining chair = K80.00
Profit on a garden chair = K50.00
The total profit $P$ is given by:

• At $(10, 20)$: $P = 80(10) + 50(20) = 800 + 1000 = 1800$
• At $(10, 30)$: $P = 80(10) + 50(30) = 800 + 1500 = 2300$
• At $(20, 60)$: $P = 80(20) + 50(60) = 1600 + 3000 = 4600$
• At $(60, 20)$: $P = 80(60) + 50(20) = 4800 + 1000 = 5800$

The maximum profit occurs at $(60, 20)$.
Mipando should make $\boxed{60 \text{ dining chairs and } 20 \text{ garden chairs}}$.

The maximum profit is the highest value calculated from the vertices.
The maximum profit is $\boxed{\text{K}5800.00}$.