Optimization Problem Modeler

Math Calculus • Applications of Derivatives

Written by STEM Calculators Team Published December 23, 2025 Updated February 24, 2026

5. Optimization Problem Modeler

Model a single-variable optimization problem, find critical points, and compare endpoints to locate the best value. Includes a zoomable/pannable objective graph with a best-point marker.

Inputs

Template

Pick a model, or use a custom objective.

Goal

Choose max or min on the interval.

Output format

LaTeX is best for readability.

Objective \(f(x)\)

Supported: + − * / ^, parentheses, variable x, constants pi, e, sin cos tan, ln log(base 10), sqrt abs exp. Implicit multiplication: 2x, (x+1)(x-1), 2sin(x).

Interval start \(a\)

Interval end \(b\)

Parameter (r / L / s)

Used by templates. Ignored for custom.

Graph center \(c\)

Auto-fit recommended.

Graph window half-width \(w\)

Graph uses \(x\in[c-w,c+w]\).

Options Show grid Mark critical points Mark endpoints Auto-fit view on solve

Ready

Graph

Drag to pan • wheel/pinch to zoom • objective curve + best point marker.

x: 0, y: 0, zoom(px/unit): 60

Result

Enter an objective and interval, then click Solve.

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5. Optimization Problem Modeler — Theory

Many optimization problems reduce to maximizing or minimizing a single-variable function \(f(x)\) on a closed interval \([a,b]\). The standard calculus method is:

\[ \textbf{(1) Find critical points:}\quad f'(x)=0 \ \text{or}\ f'(x)\ \text{does not exist (inside }(a,b)\text{)}. \] \[ \textbf{(2) Evaluate } f(x) \textbf{ at all candidates (critical points + endpoints).} \] \[ \textbf{(3) Pick the largest value (maximum) or smallest value (minimum).} \]

Why endpoints matter

Even if \(f'(x)=0\) has solutions, the global maximum/minimum on \([a,b]\) can occur at \(a\) or \(b\). This is why we compare all candidates.

Template example: rectangle in a semicircle

Suppose a rectangle is inscribed in a semicircle of radius \(r\). Let \(x\ge 0\) be half the base. Then the height is \(y=\sqrt{r^2-x^2}\), so the area is

\[ A(x)=2x\sqrt{r^2-x^2},\qquad x\in[0,r]. \]

Differentiate \(A(x)\), solve \(A'(x)=0\), then compare with endpoints \(x=0\) and \(x=r\). The optimizer is typically in the interior.

Interpreting the graph

Critical points are where the slope is zero (tangent horizontal) or undefined inside the interval.
Endpoints are included because \([a,b]\) is closed.
The best point is the candidate with the highest (or lowest) objective value, depending on the goal.

Common pitfalls

Domain restrictions (e.g., square roots require the inside \(\ge 0\)).
Objective undefined at some points (discontinuities), which must be excluded or handled carefully.
Using only \(f'(x)=0\) and forgetting endpoints.

Frequently Asked Questions

How do you solve an optimization problem on a closed interval [a,b]?

Compute critical points inside (a,b) where f'(x)=0 or f'(x) is undefined, then evaluate f(x) at each critical point and at the endpoints a and b. The largest value gives the maximum, and the smallest value gives the minimum.

Why do endpoints matter in optimization problems?

A global maximum or minimum on a closed interval can occur at the boundary even if there are interior critical points. Checking only f'(x)=0 can miss the best value at a or b.

What does it mean if the optimizer is at a critical point?

It means the best value occurs where the slope is zero (horizontal tangent) or the derivative does not exist, provided that point is in the domain and within the interval. The calculator confirms this by comparing objective values at all candidates.

What are common domain issues when modeling objectives like A(x)=2x*sqrt(r^2-x^2)?

Square roots require the inside to be nonnegative, so r^2-x^2 >= 0 restricts x to [-r,r] (often [0,r] for a geometric model). If the objective is undefined at some x, those values must be excluded from the candidate set.