Q44. - Algebra 2

Question

A feasible region has vertices at $\left(0,0\right),\left(4,0\right),\left(5,5\right),$and $\left(0,8\right)$.

Find the maximum and minimum of the function $f\left(x,y\right)=x+3y$over this region.

$\begin{array}{c}\left(A\right)\text{ maximum: }f(0,8)=24\\ \text{ minimum: }f(0,0)=0\\ \left(B\right)\text{ minimum: }f(0,0)=0\\ \text{ maximum: }f(5,5)=20\\ \left(C\right)\text{ maximum: }f(5,5)=20\\ \text{ minimum: }f(0,8)=8\\ \left(D\right)\text{ minimum: }f(4,0)=4\\ \text{ maximum: }f(0,0)=0\end{array}$

Step-by-Step Solution

Answer

The correct option is A, because, the maximum of the function $f\left(x,y\right)=x+3y$ is $f\left(0,8\right)=24,$and the minimum of the function $f\left(x,y\right)=x+3y$ is $f\left(0,0\right)=0$.

Step 1 – Evaluate the given function f x , y = x + 3 y at the given points.

Find $f\left(x,y\right)=x+3y$ at $\left(0,0\right)$:

$\begin{array}{c}f(0,0)=0+3\left(0\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}substitute }x=0,y=0\\ =0+0\\ =0\end{array}$

Find $f\left(x,y\right)=x+3y$ at $\left(4,0\right)$:

$\begin{array}{c}f(4,0)=4+3\left(0\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}substitute }x=4,y=0\\ =4+0\\ =4\end{array}$

Find $f\left(x,y\right)=x+3y$ at $\left(5,5\right)$:

$\begin{array}{c}f(5,5)=5+3\left(5\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}substitute }x=5,y=5\\ =5+15\\ =20\end{array}$

Find $f\left(x,y\right)=x+3y$ at $\left(0,8\right)$:

$\begin{array}{c}f(0,8)=0+3\left(8\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}substitute }x=5,y=5\\ =0+24\\ =24\end{array}$

So, the maximum value of $f\left(x,y\right)=x+3y$ is $f\left(0,8\right)=24$ and the minimum value of $f\left(x,y\right)=x+3y$ is $f\left(0,0\right)=0.$

Step 2 – Check whether the option A is correct or incorrect.

Here, the maximum value of $f\left(x,y\right)=x+3y$ is $f\left(0,8\right)=24$and the minimum value of $f\left(x,y\right)=x+3y$ is $f\left(0,0\right)=0.$

The above results match with the option A.

So, the option A is correct.

Step 3 – Check whether the option B is correct or incorrect.

Here, the maximum value of $f\left(x,y\right)=x+3y$ is $f\left(0,8\right)=24$and the minimum value of $f\left(x,y\right)=x+3y$ is $f\left(0,0\right)=0.$

The above results do not match with the option B.

So, the option B is incorrect.

Step 4 – Check whether the option C is correct or incorrect.

Here, the maximum value of $f\left(x,y\right)=x+3y$ is $f\left(0,8\right)=24$and the minimum value of $f\left(x,y\right)=x+3y$is $f\left(0,0\right)=0.$

 The above results do not match with the option C.

So, the option C is incorrect.

Step 5 – Check whether the option D is correct or incorrect.

Here, the maximum value of $f\left(x,y\right)=x+3y$ is $f\left(0,8\right)=24$and the minimum value of $f\left(x,y\right)=x+3y$ is $f\left(0,0\right)=0.$

 The above results do not match with the option D.

So, the option D is incorrect.

The correct option is A, because the maximum value of $f\left(x,y\right)=x+3y$ is $f\left(0,8\right)=24$and the minimum value of $f\left(x,y\right)=x+3y$ is $f\left(0,0\right)=0.$