10x	 + 	5y	≤	50
5x	 + 	10y	≤	50
5x	 + 	5y	≤	30
x	≥ 0, y ≥ 0
Find the coordinates of all corner points (if any). (Order your answers from smallest to largest x, then from smallest to largest y. If an answer does not exist, enter DNE.)


To find the coordinates of all corner points, we need to solve the system of inequalities and find the points where the boundary lines intersect.

### 1. Convert inequalities to equations:
1. \( 10x + 5y = 50 \) (Line 1)
2. \( 5x + 10y = 50 \) (Line 2)
3. \( 5x + 5y = 30 \) (Line 3)
4. \( x \geq 0, y \geq 0 \) (Constraints)

### 2. Solve the system by finding intersections of these lines:

#### Find intersection of Line 1 and Line 2:
1. Line 1: \( 10x + 5y = 50 \) simplifies to \( 2x + y = 10 \).
2. Line 2: \( 5x + 10y = 50 \) simplifies to \( x + 2y = 10 \).

Now, solve the system:
\[
2x + y = 10 \quad \text{(Equation 1)}
\]
\[
x + 2y = 10 \quad \text{(Equation 2)}
\]

From Equation 1, express \( y \) in terms of \( x \):
\[
y = 10 - 2x
\]

Substitute this into Equation 2:
\[
x + 2(10 - 2x) = 10
\]
\[
x + 20 - 4x = 10
\]
\[
-3x = -10 \quad \Rightarrow \quad x = \frac{10}{3}
\]

Now, substitute \( x = \frac{10}{3} \) into \( y = 10 - 2x \):
\[
y = 10 - 2\left(\frac{10}{3}\right) = 10 - \frac{20}{3} = \frac{30}{3} - \frac{20}{3} = \frac{10}{3}
\]

So, the intersection point of Line 1 and Line 2 is \( \left( \frac{10}{3}, \frac{10}{3} \right) \).

#### Find intersection of Line 1 and Line 3:
1. Line 1: \( 2x + y = 10 \)
2. Line 3: \( x + y = 6 \)

Now, solve the system:
\[
2x + y = 10 \quad \text{(Equation 1)}
\]
\[
x + y = 6 \quad \text{(Equation 2)}
\]

From Equation 2, express \( y \) in terms of \( x \):
\[
y = 6 - x
\]

Substitute this into Equation 1:
\[
2x + (6 - x) = 10
\]
\[
2x + 6 - x = 10
\]
\[
x = 4
\]

Now, substitute \( x = 4 \) into \( y = 6 - x \):
\[
y = 6 - 4 = 2
\]

So, the intersection point of Line 1 and Line 3 is \( (4, 2) \).

#### Find intersection of Line 2 and Line 3:
1. Line 2: \( x + 2y = 10 \)
2. Line 3: \( x + y = 6 \)

Now, solve the system:
\[
x + 2y = 10 \quad \text{(Equation 1)}
\]
\[
x + y = 6 \quad \text{(Equation 2)}
\]

From Equation 2, express \( x \) in terms of \( y \):
\[
x = 6 - y
\]

Substitute this into Equation 1:
\[
(6 - y) + 2y = 10
\]
\[
6 + y = 10
\]
\[
y = 4
\]

Now, substitute \( y = 4 \) into \( x = 6 - y \):
\[
x = 6 - 4 = 2
\]

So, the intersection point of Line 2 and Line 3 is \( (2, 4) \).

### 3. List corner points:
We now have the intersection points:
1. \( \left( \frac{10}{3}, \frac{10}{3} \right) \)
2. \( (4, 2) \)
3. \( (2, 4) \)

The corner points are ordered by increasing \( x \)-values:
1. \( (2, 4) \)
2. \( \left( \frac{10}{3}, \frac{10}{3} \right) \)
3. \( (4, 2) \)

These are the coordinates of all the corner points.

Would you like more details on any part of the solution? 

Here are some related questions:
1. How do we determine if these points lie within the feasible region?
2. Can corner points be fractions, and why are they significant in optimization problems?
3. How would you graph the system of inequalities?
4. What role does the non-negative constraint \( x \geq 0, y \geq 0 \) play in the solution?
5. What are the next steps after finding corner points for optimization?

**Tip:** Corner points are essential when solving linear programming problems since the optimal solution typically occurs at one of these points.

10x + 5y ≤ 50
5x + 10y ≤ 50
5x + 5y ≤ 30
x ≥ 0, y ≥ 0
Find the coordinates of all corner points (if any). (Order your answers from smallest to largest x, then from smallest to largest y. If an answer does not exist, enter DNE.)

Solve a system of linear inequalities to find the coordinates of corner points where lines intersect. This problem involves solving linear systems, understanding feasible regions, and applying the corner point theorem in linear programming. Suitable for high school students in Grades 10-12.

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