what is the answer to this question it is called (Solve Linear System Graphically (Lev. 1)

x≥ 0 , everthing to the right of the y-axis
y≥ 0 , everthing above the x-axis
So far you are in quadrant I of the x-y plane
y ≤ 3 , everthing from the above, but also below the horizontal line y = 3
lastly y ≤ -x + 5
sketch a line leaning 45° to the left with a y-intercept of 5, shade in everthing below that line
My diagram has a simple trapezoid consisting of a
rectangle with a right-angled triangle attached
now our Objective Function is
C = -5x + 3y , which is a line with slope 5/3
letting that line go through the origin, let it slide parallel to itself over the shaded region until you reach the farthest point from the origin.
On my diagram that looks like (5,0)