Question

The point which provides the solution of the linear programming problem, maximize $z=45x+55y$ subject to constraints $x,y\ge 0$, $6x+4y\le 120$ and $3x+10y\le 180$ is

• A. $\left(15,10\right)$
• B. $\left(10,15\right)$
• C. $\left(0,18\right)$
• D. $\left(20,0\right)$

Answer 1

The correct option is B

$\left(10,15\right)$

Explanation for the correct option:

Step 1: Find the critical points graphically

In the question, Four inequalities $x,y\ge 0$, $6x+4y\le 120$ and $3x+10y\le 180$ is given.

Draw the graph of the given inequalities.

From the graph, it is clear that the critical points are $(10,15),(20,0),(0,0)$ and $(0,18)$.

Step 2: Find the maximum value of the given function

A function $z=45x+55y$ is given in the question.

Compute the values of the given function at critical points as follows:

For $(x,y)=(10,15)$.

$$\begin{gathered} z=45\left(10\right)+55\left(15\right) \\ \Rightarrow z=450+825 \\ \Rightarrow z=1275 \end{gathered}$$

So, the value of the given function is $1275$ at $(10,15)$.

For $(x,y)=(20,0)$.

$$\begin{gathered} z=45\left(20\right)+55\left(0\right) \\ \Rightarrow z=900 \end{gathered}$$

So, the value of the given function is $900$ at $(20,0)$.

For $(x,y)=(0,0)$.

$$\begin{gathered} z=45\left(0\right)+55\left(0\right) \\ \Rightarrow z=0 \end{gathered}$$

So, the value of the given function is $0$ at $(0,0)$.

For $(x,y)=(0,18)$.

$$\begin{gathered} z=45\left(0\right)+55\left(18\right) \\ \Rightarrow z=990 \end{gathered}$$

So, the value of the given function is $990$ at $(0,18)$.

Since the maximum value of the given function is $1275$ at $(10,15)$.

Therefore, The given function maximizes at $(10,15)$.

Hence, option $B$ is the correct answer.