Chapter 11, Problem 23CT

To determine

To graph: The system of inequalities, $\{\begin{array}{l}{x}^2+y^2\le 100\\ 4x-3y\ge 0\end{array}$.

Explanation of Solution

Given information:

The system of inequalities $\{\begin{array}{l}{x}^2+y^2\le 100\\ 4x-3y\ge 0\end{array}$.

Graph:

To plot the graph of an inequality, replace inequality sign by equal sign and find the boundary curve.

$\Rightarrow \{\begin{array}{l}{x}^2+y^2=100\\ 4x-3y=0\end{array}$

If inequality is of type $\le ,\ge$ then boundary curve would be solid, if inequality is of type $<,>$ then boundary curve would be dashed curve.

Here in $x^2+y^2\le 10$, $4x-3y\ge 0$ inequality sign is $\le$ so boundary curve would be solid.

Equation $x^2+y^2=100$ represents a circle.

Comparing it with a standard form of an equation of a circle with center at origin $x^2+y^2=r^2$.

$r^2=100\Rightarrow r=10$.

So, $x^2+y^2=100$ represent a circle center at origin with radius $10$.

Now, $4x-3y=0$ represents an equation of line.

Find some ordered pairs of $4x-3y=0$.

For that choose $x$ values and calculate corresponding $y$ values. To draw a line at least two points to be known.

Choose $x=0$ and plug in $4x-3y=0$ gives $0-3y=0\Rightarrow$ $y=0$

Choose $x=3$ and plug in $4x-3y=0$ gives $12-3y=0\Rightarrow y=\frac{12}{3}=4$ $y=0$

So, the two points on a line $4x-3y=0$ are $\left(0,0\right),\left(3,4\right)$.

Now, plot ordered pairs $\left(0,0\right),\left(3,4\right)$ on a graph and by connecting them to draw a line $4x-3y=0$ and an equation of a circle $x^2+y^2=100$ with center at origin and radius $10$.

Now, to plot the graph of system of inequality $\{\begin{array}{l}{x}^2+y^2\le 100\\ 4x-3y\ge 0\end{array}$,

Choose a test point and substitute in inequality, if test point satisfies the inequality shade the region which includes test point otherwise shade the region which does not include test point.

For a line $4x-3y\ge 0$, consider $\left(1,1\right)$ as test point.

Substitute $x=1,y=1$ in $4x-3y\ge 0$

$\Rightarrow 4-3\ge 0$ True, test point satisfies the inequality. So, shade the region which includes test point $\left(1,1\right)$.

For a circle $x^2+y^2\le 100$, consider $\left(1,1\right)$ as test point.

Substitute $x=1,y=1$ in $x^2+y^2\le 100$.

$\Rightarrow 2\le 100$ True, test point satisfies the inequality. So, shade the region which includes test point $\left(1,1\right)$ that means inside a circle.

Therefore, The graph of system of inequality $\{\begin{array}{l}{x}^2+y^2\le 100\\ 4x-3y\ge 0\end{array}$ is:

Thus the intersection of the shaded region gives the resulting graph as,

Interpretation:

The graph represents a system of inequalities, $\{\begin{array}{l}{x}^2+y^2\le 100\\ 4x-3y\ge 0\end{array}$.