Answer:
[tex]35[/tex]
Step-by-step explanation:
The constraints are
The red line represents the function
[tex]y\leq \dfrac{1}{3}x+1[/tex]
At [tex]y=0[/tex]
[tex]0=\dfrac{1}{3}x+1\\\Rightarrow -1=\dfrac{1}{3}x\\\Rightarrow x=-3[/tex]
At [tex]x=0[/tex]
[tex]y=0+1\\\Rightarrow y=1[/tex]
Two points are [tex](-3,0),(0,1)[/tex]
The blue line represents the function
[tex]5\geq y+x[/tex]
at [tex]y=0[/tex]
[tex]5=x[/tex]
at [tex]x=0[/tex]
[tex]y=5[/tex]
Two points are [tex](5,0),(0,5)[/tex]
The other two constraints are [tex]x\geq 0[/tex], [tex]y\geq 0[/tex]. So, the point has to be in the first quadrant
From the graph it can be seen there are two points where the function will be maximum let us check them.
[tex](3,2)[/tex]
[tex]7x-2y=7\times 3-2\times 2=17[/tex]
[tex](5,0)[/tex]
[tex]7x-2y=7\times 5-2\times 0=35[/tex]
So, the maximum value of the function is [tex]35[/tex].
