Function value change with constrained arguments

Question

Suppose I have a function:

$$f(a,b)=h(a)+g(b)$$

where $a+b=1$. I'm interested in the values of $a$ and $b$ that maximize $f(a,b)$. Without substitution (e.g. by using the fact that $b=1-a$), how could I find out whether $f(a,b)$ is increasing in $a$?

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My approach is as follows: since an increase in $a$ leads to a corresponding decrease in $b$, there are two effects of increasing $a$:

$$\frac{dh(a)}{da};-\frac{dg(b)}{db}$$

Such that the net effect is positive iff

$$\frac{dh(a)}{da}-\frac{dg(b)}{db}>0$$

Am I correct?

Answer 1

If $(a,b)$ is a feasible point then the feasible points nearby are $(a+t,b-t)$ with $|t|\ll1$, and we have $$\eqalign{f(a+t,b-t)-f(a,b)&=\bigl(h(a+t)-h(a)\bigr)+\bigl(g(b-t)-g(b)\bigr)\cr &=t\>\bigl(h'(a)-g'(b)+o(1)\bigr)\> \qquad(t\to0) .\cr}$$ If $h'(a)-g'(b)\ne0$ then we can make the RHS positive with a suitable choice of $t$, hence $f$ cannot be maximal at the point $(a,b)$. It follows that $h'(a)=g'(b)$ is a necessary condition for a (conditional) local maximum of $f$ at $(a,b)$.

The above argument identifies the conditionally stationary points. Wether we have a minimum, maximum, or neither, in such a point has to be analyzed separately, e.g., by computing second derivatives, or through "global arguments", like convexity, values at the boundary points, etc.

Answer 2

You can find the restricted local extrema with Lagrange multipliers:

The local extrema of $$f(x,y) = h(x) + g(y)$$ with the restriction $$k(x,y) = x + y = 1$$ are given by $$\nabla f(x,y) = \lambda\nabla k(x,y).$$ In the case: $$(h'(x),g'(y)) = \lambda(1,1),$$ i.e., $$h'(x) = g'(y).$$ About your second question, the derivative of $$x\longmapsto h(x) + g(1 - x)$$ is $$h'(x) - g'(1 - x).$$ Then if $$h'(x) - g'(y) > 0,\qquad x + y = 1$$ it will be increasing, but the iff is false as the Henry's comment proves.

Answer 3

More generally, let $a$ and $b$ be functions of $t$ ($t$ could be $a$ or $b$ or something else entirely). Then $$ \frac{\mathrm{d}}{\mathrm{d}t}f(a,b)=h'(a)\frac{\mathrm{d}a}{\mathrm{d}t}+g'(b)\frac{\mathrm{d}b}{\mathrm{d}t} $$ Since $a+b=1$, we have $\frac{\mathrm{d}b}{\mathrm{d}t}=-\frac{\mathrm{d}a}{\mathrm{d}t}$. Therefore, $$ \frac{\mathrm{d}}{\mathrm{d}t}f(a,b)=(h'(a)-g'(b))\frac{\mathrm{d}a}{\mathrm{d}t} $$ This means that if $a$ is increasing, so that $\frac{\mathrm{d}a}{\mathrm{d}t}\gt0$, $f(a,b)$ will be increasing if $$ h'(a)-g'(b)\gt0 $$ So your approach appears reasonable.