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Background

I am looking for some help with the reasoning here for a few things in my game theory book (my math expertise is quite weak - below is from the text):

From Text Book

\begin{eqnarray} min \ u (x, y) = min (4xy-2x-y+3) = min (y(4x-1)-2x+3) \end{eqnarray}

(I cant get stack exchange to accept my other latex so here is a picture)

ex

graph

For each fixed x, this is a linear function in y, and therefore the point at which the minimum is attained is determined by the slope $4x-1$:

if the slope is positive the function is increasing and the minimum is attained at $y=0$;

if the slope is negative this is a decreasing function and the minimum is attained at $y=1$;

if the slope is 0, the function is constant in y and every point is a minimum point. This function of x attains a unique maximum at x= $\frac{1}{4}$, and its value there is 2.5.

What I Don't Get

What calculation is being done to determine that a minimum is attained at $y=0$ if the slope is positive, and that the minimum is attained at $y=1$ if slope is negative?

In the graph picture (linked above), I do not see how the line drawn from the peak max point ($y=2.5, x=1/4$) goes to the point $y=1, x=1$. I understand how the graph starts from the y intercept at 2 (it is the equation y=2x+2), but I don't see how it continues from there with the available data.

Thanks for your time and consideration!

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Let me give it a shot. Since for a given $x$, the function is now linear in $y$, it is either sloping up or down, and so attains the minimum at the boundary of the set of possible values $[0,1]$. The old trick of setting a derivative equal to zero doesn't work because the derivative isn't zero, it is constant on the set of possible values of $y.$ Of course, in most calculus classes you don't get many problems where you have to check the boundary points, so it's easy to forget.

Next, with respect to the graph. The graph shows the minimum over both $x$ and $y$, so you have to plug in the value of $x$ on the x-axis and then take the minimum with respect to $y$. For example, consider $x=1/8$. We have $min [y(4*1/8-1)-2x+3]$ That reduces to $min[-(1/2)y -1/4 +3]$, which is minimized when $y=1$ (because the line slopes down, and $y$ is restricted to $[0,1]$), and when $y=1$, the function is 2 1/4, which is on the line between 2 and 2 1/2 on the upward sloping part of the graph. Once $x$ exceeds 1/4, the slope turns negative and the $y$ values perforce get smaller.

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