Question 1155269
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Note that the problem says "Formulate the problem mathematically."<br>
It does not say anything about solving the problem....<br>
Assuming the problem IS supposed to be solved, here is a shortcut you can use for most problems like this where there are constraints on only two variables.<br>
(1) Plot the two major constraint boundary lines.
(2) Determine the slope of each boundary line and the slope of the constraint line.<br>
The point of the feasibility region where the objective function is maximized is where a line with the slope of the objective function just touches the feasibility region.  That point can be determine by comparing the three slopes.<br>
For this problem, the major constraint boundary lines are<br>
(1) {{{3x+2y = 150}}} --> {{{y = (-3/2)x+75}}}
(2) {{{x+2y = 75}}} --> {{{y = (-1/2)x+37.5}}}<br>
The objective function is {{{P = 20x+16y}}} which has a slope of -5/4.<br>
Because -5/4 is between -1/2 and -3/2, the point where the objective function is maximized is where the two major constraint lines intersect.<br>
So there is no need to evaluate the objective function at any point other than that point of intersection.<br>
Algebra shows the point of intersection to be (x,y) = (37.5,18.75).<br>
Note the non-integer values are probably okay if we consider the given data to be average daily production -- so that, for example, the maximum profit over 4 days is if 150 tables and 75 chairs are produced.<br>
Note also that we need to make sure the solution we found satisfies the other constraints that, up to this point, we have ignored.<br>
They do, so we have our answer.<br>
ANSWER: 37.5 tables and 18.75 chairs (per day?) for a maximum (daily) profit of 20(37.5)+16(18.75) = 750+300 = 1050 USD.<br>