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Systems of Inequalities: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A shaded half-plane

The figure shows a shaded half-plane. Write its defining inequality.

The boundary and its shaded side.

Text description of this figure

A grid with the horizontal axis labeled x and the vertical axis labeled y, both from -3 to 3, gridlines and number labels at every whole number, and the origin labeled 0. A dashed line runs through the labeled points (0, 1) and (1, -1), extending to the edges of the grid. The side of the line containing the origin is shaded; the origin itself is not marked. No inequality or equation is written on the figure.
1. Hint 1
  Use two points on its boundary to find the line.
2. Hint 2
  Use an interior point and the boundary style to choose the symbol.
Answer

$y<-2x+1$ , or $2x+y<1$ .

Full solution

The boundary passes through $(0,1)$ and $(1,-1)$ , giving slope $-2$ and line $y=-2x+1$ .
The shaded point $(0,0)$ lies below this line.
The line is dashed, so its points are excluded.
$y<-2x+1$

Answer

$y<-2x+1$ , or $2x+y<1$ .

Key idea

A boundary equation, a shaded side, and inclusion information determine a half-plane.
Problem 1: Mark complete
Problem 2 A boundary choice

On the blank coordinate grid in the figure, graph $x\ge-1$ and $y<2$ together.

A blank coordinate grid for the two boundaries.

Text description of this figure

A grid with the horizontal axis labeled x running from -4 to 4 and the vertical axis labeled y running from -4 to 4, gridlines and number labels at every whole number, and the origin labeled 0. No point, boundary, or shading is drawn.
1. Hint 1
  Each condition keeps one side of a horizontal or vertical line.
2. Hint 2
  The equality signs decide which of the two boundaries is included.
Answer

The overlap is right of the solid line $x=-1$ and below the dashed line $y=2$ .

Full solution

The condition $x\ge-1$ includes its vertical boundary and all points to its right.
The condition $y<2$ excludes its horizontal boundary and keeps the lower side.
A point such as $(0,0)$ passes both tests.
Shade their overlap.

Answer

The overlap is right of the solid line $x=-1$ and below the dashed line $y=2$ .

Key idea

A system keeps only points accepted by every inequality.
Problem 2: Mark complete
Problem 3 A fixed horizontal slice

For the system $y\ge x-2$ and $y\le5-x$ , determine all real $x$ allowed on the horizontal line $y=1$ .
1. Hint 1
  The chosen height turns both conditions into one-variable inequalities.
2. Hint 2
  Find the overlap of the resulting upper bounds on $x$ .
Answer

$x\le3$ , or $(-\infty,3]$ .

Full solution

Substitute $y=1$ in both conditions.
$\begin{aligned} 1&\ge x-2 \\ 1&\le5-x \end{aligned}$

These require $x\le3$ and $x\le4$ .
Their overlap is $x\le3$ .
At $x=3$ , both original inequalities hold.

Answer

$x\le3$ , or $(-\infty,3]$ .

Key idea

A horizontal slice of a feasible region is found by fixing its height in every condition.
Problem 3: Mark complete
Problem 4 A bounded patch

On the blank grid in the figure, graph $x\ge1$ , $y\ge0$ , $y\le x$ , and $x+y\le6$ . Give every corner of the feasible region.

A blank coordinate grid for the region.

Text description of this figure

A grid with the horizontal axis labeled x running from -1 to 7 and the vertical axis labeled y running from -1 to 7, gridlines and number labels at every whole number, and the origin labeled 0. No point, boundary, or shading is drawn.
1. Hint 1
  Intersect all four allowed sides before keeping a boundary crossing.
2. Hint 2
  Check each crossing against the inequalities it did not come from.
Answer

Corners: $(1,0)$ , $(1,1)$ , $(3,3)$ , $(6,0)$ ; all edges solid.

Full solution

The lines $x=1$ and $y=0$ give $(1,0)$ .
Intersecting $x=1$ and $y=x$ gives $(1,1)$ .
For the slanted pair,
$\begin{aligned} x+y&=6 \\ y&=x \\ 2x&=6 \end{aligned}$

This gives $(3,3)$ .
Finally $y=0$ and $x+y=6$ give $(6,0)$ .
All pass all four conditions.
The crossing $(1,5)$ fails $y\le x$ and is discarded, and the crossing $(0,0)$ from $y=0$ and $y=x$ fails $x\ge1$ and is also discarded.
Every boundary includes equality, so all edges are solid.

Answer

Corners: $(1,0)$ , $(1,1)$ , $(3,3)$ , $(6,0)$ ; all edges solid.

Key idea

A boundary crossing is a corner only if it satisfies the whole system.
Problem 4: Mark complete
Problem 5 A cooling schedule

A cooling system runs two modes for real durations $x$ and $y$ hours. It requires $x\ge1$ , $y\ge1$ , $x+y\le5$ , and $x\le3$ . Cooling output is $C=2x+5y$ units. Find the greatest output and the durations achieving it.
1. Hint 1
  The linear output can be evaluated at the corners of the allowed region.
2. Hint 2
  Find the intersections of the four boundaries and keep the feasible ones.
Answer

Maximum 22 units at $x=1$ hour and $y=4$ hours.

Full solution

Intersect the boundaries in pairs. $x=1$ with $y=1$ gives $(1,1)$ ; $x=1$ with $x+y=5$ gives $(1,4)$ ; $x=3$ with $y=1$ gives $(3,1)$ ; $x=3$ with $x+y=5$ gives $(3,2)$ .
The remaining pair, $y=1$ and $x+y=5$ , gives $(4,1)$ , which fails $x\le3$ and is discarded.
The region is bounded with solid boundaries.
Evaluate each of the four kept corners.
$\begin{aligned} C(1,1)&=7 \\ C(3,1)&=11 \\ C(3,2)&=16 \\ C(1,4)&=22 \end{aligned}$

The largest output is 22 units.
The schedule $(1,4)$ satisfies all four conditions.

Answer

Maximum 22 units at $x=1$ hour and $y=4$ hours.

Key idea

A linear objective on a bounded region with solid boundaries reaches its best value at a corner.
Problem 5: Mark complete
Problem 6 A slanted strip

Find the minimum and maximum of $P=x-2y$ subject to $0\le x\le4$ and $x\le y\le x+2$ . State where each is attained.
1. Hint 1
  An extremum of a linear expression over a bounded region with solid boundaries occurs at one of its corners.
2. Hint 2
  Find the region's four corners, then evaluate $P$ at each.
Answer

Minimum $-8$ at $(4,6)$ ; maximum $0$ at $(0,0)$ .

Full solution

The region is a parallelogram cut by $x=0$ and $x=4$ : at $x=0$ the strip runs from $y=0$ to $y=2$ , and at $x=4$ from $y=4$ to $y=6$ , giving the four corners $(0,0)$ , $(0,2)$ , $(4,4)$ , $(4,6)$ .
$\begin{aligned} P(0,0)&=0 \\ P(0,2)&=-4 \\ P(4,4)&=-4 \\ P(4,6)&=-8 \end{aligned}$

The smallest value is $-8$ and the largest is $0$ .
Both corner points satisfy all the constraints.

Answer

Minimum $-8$ at $(4,6)$ ; maximum $0$ at $(0,0)$ .

Key idea

A strip cut off at two inputs can form a bounded region with both extrema.
Problem 6: Mark complete
Problem 7 Resource requirements

A process requires real amounts $x,y\ge0$ , together with $x+2y\ge8$ , $x\le2$ , and $y\le2$ . Determine whether any allowed amounts exist and whether $x+y$ has an optimum there.
1. Hint 1
  Use the upper bounds to estimate the largest possible left side of the requirement.
2. Hint 2
  Compare that bound with the required minimum.
Answer

The region is empty; no minimum or maximum exists.

Full solution

The upper bounds imply
$x+2y\le6$
This contradicts $x+2y\ge8$ .
Thus no point is allowed, and there is nowhere to evaluate an objective.

Answer

The region is empty; no minimum or maximum exists.

Key idea

Separate upper bounds can make a combined minimum requirement impossible.
Problem 7: Mark complete
Problem 8 Points between two choices

Two points $A$ and $B$ satisfy $2x+y\le9$ . A student claims their midpoint also satisfies it. Is this correct? Justify without assuming where the points lie.
1. Hint 1
  The midpoint averages the two coordinates.
2. Hint 2
  Evaluate the linear expression at the averaged coordinates.
Answer

Yes; the midpoint satisfies the inequality.

Full solution

Write $A=(a,b)$ and $B=(c,d)$ ; the hypotheses are $2a+b\le9$ and $2c+d\le9$ .
Their midpoint is $\left(\frac{a+c}2,\frac{b+d}2\right)$ , so twice its first coordinate plus its second coordinate is
$\begin{aligned} S&=2\cdot\frac{a+c}2+\frac{b+d}2 \\ &=(a+c)+\frac{b+d}2 \\ &=\frac{(2a+b)+(2c+d)}2 \end{aligned}$
The numerator is at most $18$ , so the value is at most $9$ .

Answer

Yes; the midpoint satisfies the inequality.

Key idea

Averaging two accepted points preserves a linear inequality.
Problem 8: Mark complete
Problem 9 A claimed maximum

For $0\le x\le2$ and $0\le y<3$ , a student says $P=x+y$ has a maximum of $5$ at $(2,3)$ . Is this correct? Explain.
1. Hint 1
  Check whether the proposed maximizing point belongs to the region.
2. Hint 2
  For any allowed height, consider a slightly larger height still below the boundary.
Answer

No; no maximum exists, although $P$ approaches $5$ .

Full solution

The point $(2,3)$ is excluded because $y<3$ .
Every allowed point has $x+y<5$ .
At $x=2$ , values of $y$ can approach $3$ from below, so $P$ approaches $5$ .
For any allowed $y$ , the new height $(y+3)/2$ is larger and still below $3$ , increasing $P$ .

Answer

No; no maximum exists, although $P$ approaches $5$ .

Key idea

A missing boundary can prevent a bounded objective from attaining its upper bound.
Problem 9: Mark complete
Problem 10 An unbounded direction

The region is $x\ge0$ and $-1\le y\le2$ . Does $P=y$ have both a maximum and a minimum even though the region is unbounded? Explain and describe where they occur.
1. Hint 1
  Consider which coordinate direction is unbounded, and whether the objective depends on that direction.
2. Hint 2
  Check which coordinate the objective actually measures.
Answer

Yes; minimum $-1$ at $(x,-1)$ and maximum $2$ at $(x,2)$ , for all $x\ge0$ .

Full solution

The region extends indefinitely as $x$ increases.
However,
$\begin{aligned} P&\ge-1 \\ P&\le2 \end{aligned}$
Both bounding heights are included.
Every point $(x,-1)$ with $x\ge0$ attains the minimum, and every point $(x,2)$ attains the maximum.

Answer

Yes; minimum $-1$ at $(x,-1)$ and maximum $2$ at $(x,2)$ , for all $x\ge0$ .

Key idea

An unbounded region can still bound an objective that ignores its unbounded direction.
Problem 10: Mark complete

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