Hard Further Maths Linear programming (graphical) Questions

Write down the constraints

The feasible region is the set of points satisfying all of these inequalities at once.

State the objective function

The objective is the linear expression to be maximised.

Recall the vertex (extreme point) theorem

So it is enough to test the corners of the region rather than every point of it.

Solve $5x+5y=34$ and $2x-y=0$ simultaneously

This intersection satisfies every constraint, so $\left(\frac{34}{15},\frac{68}{15}\right)$ is a vertex of $R$.

Solve $5x+5y=34$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(\frac{34}{5},0\right)$ is a vertex of $R$.

Solve $3x-y=0$ and $2x-y=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

Solve $3x-y=0$ and $x=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

Solve $3x-y=0$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

Solve $2x-y=0$ and $x=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

Solve $2x-y=0$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

Solve $x=0$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

List the vertices of the feasible region

The feasible region is a convex polygon with these 3 corners.

Evaluate $P$ at $\left(0,0\right)$

The objective is evaluated by substituting the coordinates of the vertex.

Evaluate $P$ at $\left(\frac{34}{15},\frac{68}{15}\right)$

The objective is evaluated by substituting the coordinates of the vertex.

Evaluate $P$ at $\left(\frac{34}{5},0\right)$

The objective is evaluated by substituting the coordinates of the vertex.

Collect the values of the objective at every vertex

Every corner of the region has now been tested.

Compare the values and select the largest

No other vertex gives a larger value, so the optimum is unique.

Select the correct statement about the objective line

Rearranging $P=x+6y$ gives $y=-\frac{1}{6}x+\frac{P}{6}$, so every objective line has gradient $-\frac{1}{6}$.

Write down the constraints

The feasible region is the set of points satisfying all of these inequalities at once.

State the objective function

The objective is the linear expression to be maximised.

Note that $x$ and $y$ must be integers

The optimum must be an integer point of $R$, so the continuous optimum is only a starting point.

Solve $5x+5y=37$ and $x+3y=21$ simultaneously

This intersection satisfies every constraint, so $\left(\frac{3}{5},\frac{34}{5}\right)$ is a vertex of $R$.

Solve $5x+5y=37$ and $5x+y=28$ simultaneously

This intersection satisfies every constraint, so $\left(\frac{103}{20},\frac{9}{4}\right)$ is a vertex of $R$.

Solve $x+3y=21$ and $x=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,7\right)$ is a vertex of $R$.

Solve $5x+y=28$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(\frac{28}{5},0\right)$ is a vertex of $R$.

Solve $x=0$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

List the vertices of the feasible region

The feasible region is a convex polygon with these 5 corners.

Collect the values of the objective at every vertex

Every corner of the region has now been tested.

Compare the values and select the largest

No other vertex gives a larger value, so the optimum is unique.

Round the continuous optimum and test the rounded point

The rounded point is not even feasible, so rounding the continuous optimum is not a valid method.

Count the integer points of the feasible region

The region is small enough to search its integer points exhaustively.

Evaluate the objective at the best integer candidates

Searching the integer points is the only safe method; the answer is the best of them.

Select the best feasible integer point

This is the optimal integer solution.

Select the optimal integer point

Rounding the continuous optimum gives $\left(1,7\right)$, which is not the answer; only a search of the integer points is safe.

Write down the constraints

The feasible region is the set of points satisfying all of these inequalities at once.

State the objective function

The objective is the linear expression to be maximised.

Note that $x$ and $y$ must be integers

The optimum must be an integer point of $R$, so the continuous optimum is only a starting point.

Solve $2x+y=22$ and $3x-y=0$ simultaneously

This intersection satisfies every constraint, so $\left(\frac{22}{5},\frac{66}{5}\right)$ is a vertex of $R$.

Solve $2x+y=22$ and $x=7$ simultaneously

This intersection satisfies every constraint, so $\left(7,8\right)$ is a vertex of $R$.

Solve $3x-y=0$ and $x=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

Solve $3x-y=0$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

Solve $x=7$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(7,0\right)$ is a vertex of $R$.

Solve $x=0$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

List the vertices of the feasible region

The feasible region is a convex polygon with these 4 corners.

Collect the values of the objective at every vertex

Every corner of the region has now been tested.

Compare the values and select the largest

No other vertex gives a larger value, so the optimum is unique.

Round the continuous optimum and test the rounded point

The rounded point is not even feasible, so rounding the continuous optimum is not a valid method.

Count the integer points of the feasible region

The region is small enough to search its integer points exhaustively.

Evaluate the objective at the best integer candidates

Searching the integer points is the only safe method; the answer is the best of them.

Select the best feasible integer point

This is the optimal integer solution.

Select the correct statement

Rounding the continuous optimum is not a valid method: the integer points must be searched.

Write down the constraints

The feasible region is the set of points satisfying all of these inequalities at once.

State the objective function

The objective is the linear expression to be maximised.

Note that $x$ and $y$ must be integers

The optimum must be an integer point of $R$, so the continuous optimum is only a starting point.

Solve $2x+y=35$ and $y=7$ simultaneously

This intersection satisfies every constraint, so $\left(14,7\right)$ is a vertex of $R$.

Solve $2x+y=35$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(\frac{35}{2},0\right)$ is a vertex of $R$.

Solve $y=7$ and $3x-y=0$ simultaneously

This intersection satisfies every constraint, so $\left(\frac{7}{3},7\right)$ is a vertex of $R$.

Solve $3x-y=0$ and $x=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

Solve $3x-y=0$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

Solve $x=0$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

List the vertices of the feasible region

The feasible region is a convex polygon with these 4 corners.

Evaluate $P$ at $\left(0,0\right)$

The objective is evaluated by substituting the coordinates of the vertex.

Evaluate $P$ at $\left(\frac{7}{3},7\right)$

The objective is evaluated by substituting the coordinates of the vertex.

Collect the values of the objective at every vertex

Every corner of the region has now been tested.

Compare the values and select the largest

No other vertex gives a larger value, so the optimum is unique.

Round the continuous optimum and test the rounded point

The rounded point is not even feasible, so rounding the continuous optimum is not a valid method.

Count the integer points of the feasible region

The region is small enough to search its integer points exhaustively.

Evaluate the objective at the best integer candidates

Searching the integer points is the only safe method; the answer is the best of them.

Select the best feasible integer point

This is the optimal integer solution.

Select the correct statement

Rounding the continuous optimum is not a valid method: the integer points must be searched.

Write down the constraints

The feasible region is the set of points satisfying all of these inequalities at once.

State the objective function

The objective is the linear expression to be maximised.

Note that $x$ and $y$ must be integers

The optimum must be an integer point of $R$, so the continuous optimum is only a starting point.

Solve $x+4y=20$ and $2x+3y=21$ simultaneously

This intersection satisfies every constraint, so $\left(\frac{24}{5},\frac{19}{5}\right)$ is a vertex of $R$.

Solve $x+4y=20$ and $3x-y=0$ simultaneously

This intersection satisfies every constraint, so $\left(\frac{20}{13},\frac{60}{13}\right)$ is a vertex of $R$.

Solve $2x+3y=21$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(\frac{21}{2},0\right)$ is a vertex of $R$.

Solve $3x-y=0$ and $x=0$ simultaneously

This intersection satisfies every constraint, so $\left(0,0\right)$ is a vertex of $R$.

List the vertices of the feasible region

The feasible region is a convex polygon with these 4 corners.

Collect the values of the objective at every vertex

Every corner of the region has now been tested.

Compare the values and select the largest

No other vertex gives a larger value, so the optimum is unique.

Round the continuous optimum and test the rounded point

The rounded point is not even feasible, so rounding the continuous optimum is not a valid method.

Count the integer points of the feasible region

The region is small enough to search its integer points exhaustively.

Evaluate the objective at the best integer candidates

Searching the integer points is the only safe method; the answer is the best of them.

Select the best feasible integer point

This is the optimal integer solution.

State the optimal integer value

This is the best value attainable at an integer point of $R$.