Regions Between Two Curves

Introduction

A region between two curves has a simple description: for each $x$ in an interval, $y$ lies between a lower graph and an upper graph. Its boundary, however, includes more than those two curved arcs. Unless the graphs meet at both ends of the interval, vertical segments join them. An absolute-extrema search must account for every part of that boundary.

The method is to examine interior critical points, restrict the function to each boundary piece, and include the points where pieces meet. The last step is important: the restrictions on the graph arcs do not, by themselves, check the interiors of the vertical sides.

Describing the Region

Definition: Let $a<b$, and let $g$ and $h$ be continuous functions on $[a,b]$ such that $g(x)\le h(x)$ for every $x\in[a,b]$. The closed region between their graphs is $$ R=\{(x,y):a\le x\le b,\ g(x)\le y\le h(x)\}. $$ Its boundary consists of the lower graph $y=g(x)$, the upper graph $y=h(x)$, and the vertical segments at $x=a$ and $x=b$, with any coincident endpoints counted just once.

Continuity of $g$ and $h$ makes $R$ closed and bounded, hence compact. If $f$ is continuous on $R$, the extreme value theorem guarantees that $f$ attains an absolute minimum and an absolute maximum there. Finding them requires checking the possible locations where those values can occur.

Assume for the usual derivative-based search that $f$ is differentiable in the interior of $R$, and that the boundary restrictions used below are differentiable on the relevant intervals. Interior candidates satisfy $f_x=f_y=0$. On a graph boundary, substitute the graph equation into $f$, producing a one-variable function. On a vertical side, hold $x$ fixed and regard $f$ as a function of $y$.

How to Organize the Candidate Search

Find interior candidates.
Solve $f_x(x,y)=0$ and $f_y(x,y)=0$, retaining only solutions in the interior of $R$.

Restrict to the two graph arcs.
Form $F_g(x)=f(x,g(x))$ and $F_h(x)=f(x,h(x))$. Find their interior critical points and include the endpoint values at $x=a$ and $x=b$.

Restrict to the vertical sides.
On $x=a$ and $x=b$, find critical points of $f(a,y)$ and $f(b,y)$ over their respective $y$-intervals. Include the endpoints of each interval.

Compare all candidate values.
Evaluate $f$ at every candidate. The largest value is the absolute maximum, and the smallest is the absolute minimum.

The graph-arc restrictions satisfy the chain rule: $F_g'(x)=f_x(x,g(x))+f_y(x,g(x))g'(x)$, with the corresponding formula using $h$ for the upper graph. A critical point of a restriction is a candidate on that arc; it is not automatically an extremum of the original two-variable function.

Why These Checks Are Complete

The candidate method follows from the locations where an extremum can occur. Because $R$ is compact and $f$ is continuous, an absolute extremum exists. If it occurs in the interior, Fermat’s theorem says both first partial derivatives vanish. Otherwise it lies on the boundary.

The boundary is made from four pieces: the lower graph, upper graph, left vertical segment, and right vertical segment. On the interior of any one smooth piece, an extremum of $f$ restricted to that piece must be a critical point of its one-variable restriction. If it occurs at the end of a piece, it is a meeting point of boundary pieces and must be checked as an endpoint. Thus the search includes both the critical points of the restrictions and all their endpoints.

Key takeaway: Checking the two graph restrictions and their endpoints is not enough when a vertical side has positive length. Each vertical side has its own one-variable restriction and its own possible interior critical points.

Worked Example: A Curved Band and Its Full Boundary

Worked Example: Find the Absolute Extrema on a Curved Band

Find the absolute minimum and maximum of $f(x,y)=x^2+y^2$ on the region $-1\le x\le1$, $x^2\le y\le2$. The lower graph is $y=x^2$, the upper graph is $y=2$, and the vertical sides run from $y=1$ to $y=2$.

Interior. The partial derivatives are $f_x=2x$ and $f_y=2y$. Their only simultaneous zero is $(0,0)$, which lies on the lower boundary, not in the interior. There are no interior candidates.

Lower graph. Substitution gives $F_g(x)=x^2+(x^2)^2=x^2+x^4$. Its derivative is $2x+4x^3=2x(1+2x^2)$. Since $1+2x^2>0$, the only critical point is $x=0$. The lower-graph candidate values are $F_g(0)=0$, $F_g(-1)=1+1=2$, and $F_g(1)=1+1=2$.

Upper graph. Substitution gives $F_h(x)=x^2+2^2=x^2+4$, with derivative $2x$. The critical point is $x=0$, where $F_h(0)=4$. At the endpoints, $F_h(-1)=1+4=5$ and $F_h(1)=1+4=5$.

Vertical sides. At $x=-1$, $f(-1,y)=1+y^2$ for $1\le y\le2$; its derivative with respect to $y$ is $2y$, which has no zero on this interval. The endpoint values are $1+1^2=2$ and $1+2^2=5$. At $x=1$, the restriction is also $1+y^2$, giving the same endpoint values and no side-interior critical point.

Comparing the candidates gives an absolute minimum of $0$ at $(0,0)$, and an absolute maximum of $5$ at $(-1,2)$ and $(1,2)$. Substitution checks these values directly: $f(0,0)=0$, while $f(\pm1,2)=1+4=5$.

Worked Example: An Interior Minimum and Boundary Maximum

Worked Example: Optimize a Function Adapted to a Curved Band

On $R=\{(x,y):0\le x\le1,\ x^2\le y\le x^2+2\}$, find the absolute extrema of $f(x,y)=(x-\tfrac12)^2+(y-x^2-1)^2$.

Interior. Write $q=y-x^2-1$. Then $f_y=2q$ and $f_x=2(x-\tfrac12)-4xq$. Setting $f_y=0$ gives $q=0$. Substituting into $f_x=0$ gives $2(x-\tfrac12)=0$, so $x=\tfrac12$. Then $y=x^2+1=\tfrac14+1=\tfrac54$, which is in the interior of the region. The function value is $0+0=0$.

Lower graph. When $y=x^2$, $q=-1$, so the restriction is $(x-\tfrac12)^2+1$. Its derivative is $2(x-\tfrac12)$, which vanishes at $x=\tfrac12$, giving value $1$. At $x=0$ and $x=1$, the value is $\tfrac14+1=\tfrac54$.

Upper graph. When $y=x^2+2$, $q=1$, so the restriction is also $(x-\tfrac12)^2+1$. Its critical point is $x=\tfrac12$, with value $1$; its endpoint values are both $\tfrac54$.

Vertical sides. On either vertical side, $q$ ranges from $-1$ to $1$. The restriction is $\tfrac14+q^2$, whose derivative with respect to $q$ is $2q$. The side-interior critical point $q=0$ has value $\tfrac14$. At $q=-1$ and $q=1$, the value is $\tfrac14+1=\tfrac54$.

All candidate values are $0$, $\tfrac14$, $1$, or $\tfrac54$. Therefore the absolute minimum is $0$ at $(\tfrac12,\tfrac54)$, and the absolute maximum is $\tfrac54$ at the four corners. Substitution gives $f(\tfrac12,\tfrac54)=0$; at each corner, $(x-\tfrac12)^2=\tfrac14$ and $q^2=1$, so $f=\tfrac54$.

A Common Pitfall: Omitting the Vertical-Side Interior

Endpoint checks on the graph arcs include the corners, but they do not include interior points of the vertical sides. The following example shows that a side-interior point can be the absolute maximum.

Worked Example: A Maximum Between the Side Endpoints

On the rectangle $0\le x\le1,\ 0\le y\le2$, find the absolute extrema of $f(x,y)=(x-\tfrac12)^2-(y-1)^2$. This rectangle is a region between the constant graphs $y=0$ and $y=2$.

Interior. The partial derivatives are $f_x=2(x-\tfrac12)$ and $f_y=-2(y-1)$. The only interior critical point is $(\tfrac12,1)$, with value $0$.

Graph boundaries. At $y=0$ and $y=2$, the restriction is $(x-\tfrac12)^2-1$. Its derivative is $2(x-\tfrac12)$, so its critical point is $x=\tfrac12$, with value $-1$. At $x=0$ and $x=1$, the values on these graph boundaries are $\tfrac14-1=-\tfrac34$.

Vertical sides. On either side, the restriction is $\tfrac14-(y-1)^2$. Its derivative is $-2(y-1)$, which vanishes at $y=1$, giving value $\tfrac14$. At $y=0$ and $y=2$, it gives $\tfrac14-1=-\tfrac34$.

Thus the absolute minimum is $-1$ at $(\tfrac12,0)$ and $(\tfrac12,2)$, while the absolute maximum is $\tfrac14$ at $(0,1)$ and $(1,1)$. For a concrete incorrect calculation, checking only the two graph boundaries and their endpoints would report $-\tfrac34$ as the largest value among those candidates. The missing side-interior checks reveal the larger value $\tfrac14$.

Checklist

For a region between two graphs, make the boundary decomposition explicit before calculating. At each end of the $x$-interval, the vertical side runs from the lower graph’s value to the upper graph’s value. Include both ends of that side, even though they also appear as graph-arc endpoints. Repeatedly checking a corner does no harm; omitting a boundary piece can.

Check Your Understanding

Use the candidate-search method to answer these questions.

For $a\le x\le b$ and $g(x)\le y\le h(x)$, what are the four types of boundary pieces to check?
What one-variable function results from restricting $f$ to the lower graph?
On the side $x=b$, which derivative condition identifies an interior critical point of the side restriction?
Why must endpoints of each boundary piece be included even when the restriction has no critical point there?
For $f(x,y)=(x-\tfrac12)^2-(y-1)^2$ on $0\le x\le1,\ 0\le y\le2$, what side-interior points give the absolute maximum?

What You'll Learn