Do you mind elaborating by what you meant by "Being able to recognise that if all number on a line are a solution, the gradient of the objective function must be the exact same as this line. See 3d 2016 VCAA". I figured that the point (400,100) lies on the line x+y=500 but I don't understand how m and n are equal and how the gradient is -1 for the profit function, Q. Thanks.
In the simplest case, when an objective function attains its maximum value at a corner point of a feasible region, that is the
only point in the feasible region where the maximum value is attained.
In slightly more complicated cases, the objective function attains its maximum value at multiple points of the feasible region, but each one of these is a corner point.
In more interesting cases, the objective function may attain its maximum value at two corner points (and possibly others) that are
on the same line (ie. same boundary of the feasible region). When this occurs, it turns out that the objective function also attains that maximum value
at each point along this line. (The reasons for this are beyond the scope of the course).
The only way that the objective function can take on the same value at each point along a single boundary of the feasible region is if the gradient of the objective function is the same as that boundary. If its gradient were different, then there would be some point on that boundary which would not give the same value for the objective function as another point on that boundary.
For that particular question (VCAA 2016), the gradient of the line x + y = 500 is –1 (rearrange to y = –x + 500). For a profit function mx + ny = P, the gradient will be –1 if m = n, since rearraging gives y = –(m/n)x + 500/n.
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