Further Maths Linear programming (graphical) Practice Questions

State the objective function

The objective is the linear expression whose value is required.

Confirm that $\left(2,4\right)$ is a vertex by checking the boundaries through it

Two boundary lines meet at this point and it satisfies every constraint.

Substitute the coordinates of the vertex

The objective is evaluated at a point by substituting its coordinates.

Work out the value

This is the value of the objective at this corner of $R$.

Write down the constraints that must be satisfied

A point lies in $R$ only if it satisfies every one of them.

Test each point in $3x+y\le15$

Any point failing this constraint can be discarded at once.

Test the surviving points in $y\le12$

Only points that pass every test lie in the feasible region.

State which point is feasible

This is the only one of the five points satisfying all of the constraints.

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.

Solve $5x+3y=35$ and $x-y=0$ simultaneously

This intersection satisfies every constraint, so $\left(\frac{35}{8},\frac{35}{8}\right)$ is a vertex of $R$.

List the vertices of the feasible region

The feasible region is a convex polygon with these 3 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.

Select the correct statement about the objective line

Rearranging $P=9x+6y$ gives $y=-\frac{3}{2}x+\frac{P}{6}$, so every objective line has gradient $-\frac{3}{2}$.

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 minimised.

Recall the vertex (extreme point) theorem

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

Solve $4x+3y=30$ and $4x+y=26$ simultaneously

This intersection satisfies every constraint, so $\left(6,2\right)$ is a vertex of $R$.

Solve $4x+3y=30$ and $y=0$ simultaneously

This intersection satisfies every constraint, so $\left(\frac{15}{2},0\right)$ is a vertex of $R$.

Solve $4x+y=26$ and $x=2$ simultaneously

This intersection satisfies every constraint, so $\left(2,18\right)$ is a vertex of $R$.

List the vertices of the feasible region

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

Evaluate $C$ at $\left(2,18\right)$

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

Evaluate $C$ at $\left(6,2\right)$

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

Evaluate $C$ at $\left(\frac{15}{2},0\right)$

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 smallest

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

Select the correct statement about the objective line

Rearranging $C=3x+2y$ gives $y=-\frac{3}{2}x+\frac{C}{2}$, so every objective line has gradient $-\frac{3}{2}$.

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 $-\frac{1}{6}$.