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Optimization & Related Rates Lab

Explore optimization and related rates problems from AP Calculus. Visualize objective functions, find critical points with derivatives, and watch related rates change in real time with animated diagrams.

Guided Experiment: Optimization Strategy

If you follow the steps (write objective, apply constraint, differentiate, set f′ = 0, test with f″), will you always find the correct optimum?

Write your hypothesis in the Lab Report panel, then click Next.

Diagram

x

Objective Function Graph

Controls

x (half-width)0.5000
0.050.95

Objective

Maximize the area of a rectangle inscribed in a semicircle of radius 1

Constraint

The top corners touch the semicircle y = √(1 − x²)

Results

A(x)=2x1−x2A(x) = 2x\sqrt{1 - x^2}
f(x) at x = 0.5000
0.8660
f′(x)
1.1547
f″(x)
-3.8490
Classification
Not critical

Critical Point Solution

Critical x0.7071
Optimal value1.0000
f′ at critical0.000000
f″ at critical-8.0000
TypeMaximum

Step-by-Step

1
A(x)=2x1−x2A(x) = 2x\sqrt{1 - x^2}
2
A′(x)=2(1−2x2)1−x2A'(x) = \frac{2(1 - 2x^2)}{\sqrt{1 - x^2}}
3
Set A′(x)=0:1−2x2=0  ⟹  x=12\text{Set } A'(x) = 0: \quad 1 - 2x^2 = 0 \implies x = \frac{1}{\sqrt{2}}
4
A′′(12)=2⋅12(2⋅12−3)(1−12)3/2<0  ⟹  maximumA''\left(\frac{1}{\sqrt{2}}\right) = \frac{2 \cdot \frac{1}{\sqrt{2}}(2 \cdot \frac{1}{2} - 3)}{(1 - \frac{1}{2})^{3/2}} < 0 \implies \text{maximum}
5
A(12)=2⋅12⋅12=1A\left(\frac{1}{\sqrt{2}}\right) = 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = 1

Data Table

(0 rows)
#ProblemVariableCritical ValueOptimal Valuef′(x)f″(x)Max/Min
0 / 500
0 / 500
0 / 500

Reference Guide

Optimization Strategy

The general strategy for solving optimization problems has five steps.

  1. Identify the objective function to maximize or minimize
  2. Write the constraint and eliminate one variable
  3. Find the derivative and set f'(x) = 0
  4. Solve for the critical value(s)
  5. Verify with the second derivative test or endpoint check
f′(xc)=0andf′′(xc)≠0f'(x_c) = 0 \quad \text{and} \quad f''(x_c) \ne 0

First & Second Derivative Tests

The first derivative test checks whether f'(x) changes sign at a critical point.

f′(x) changes +→−  ⟹  local maxf'(x) \text{ changes } + \to - \implies \text{local max}

The second derivative test provides a quicker classification when f''(x) exists.

f′′(xc)<0  ⟹  max,f′′(xc)>0  ⟹  minf''(x_c) < 0 \implies \text{max}, \quad f''(x_c) > 0 \implies \text{min}

Related Rates

Related rates problems connect two changing quantities through a geometric or physical relationship.

  1. Draw a diagram and assign variables
  2. Write an equation relating the variables
  3. Differentiate both sides with respect to time t
  4. Substitute known values and rates
  5. Solve for the unknown rate
dydt=dydx⋅dxdt\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}

Implicit Differentiation

When variables are linked by an equation (not solved for y), differentiate implicitly.

ddt[x2+y2=L2]  ⟹  2xdxdt+2ydydt=0\frac{d}{dt}[x^2 + y^2 = L^2] \implies 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

This technique is essential for related rates. Every term with a variable gets a chain rule factor of d(var)/dt.

ddt[f(g(t))]=f′(g(t))⋅g′(t)\frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t)