Maxima, Minima & Time Rates
Maxima and Minima
Key Ideas
- First derivative zero gives critical point
- Second derivative sign indicates concavity
- Optimization equates derivative to zero
Relative Extrema are peak or trough values where the first derivative equals zero.
Conditions for Extrema and Inflection
Maximum Point Condition
Minimum Point Condition
Point of Inflection Condition
Optimization Steps
- Draw figure and identify variables
- Formulate primary equation to optimize
- Express equation in one variable
- Differentiate and equate to zero
- Solve for critical variable value
Geometric Shapes and Variable Relationships
| Formula / Relationship | Figure |
|---|---|
| Rectangle in Circle: \(x = y = r\sqrt{2}\) (square) | |
| Rectangle in Semicircle: \(b = 2h\), \(x = r\sqrt{2}\), \(y = \dfrac{r}{\sqrt{2}}\) | |
| Rectangle in Triangle: \(x = \dfrac{b}{2}\), \(y = \dfrac{h}{2}\) | |
| Rectangle in Ellipse: \(x = \dfrac{a}{\sqrt{2}}\), \(y = \dfrac{b}{\sqrt{2}}\) (\(x\) and \(y\) are the half-width and half-height; the full rectangle is \(a\sqrt{2}\) by \(b\sqrt{2}\)) | |
| Rectangle with Fixed Perimeter: \(x = y\) (square) | |
| Three-Sided Fence Lot: \(x = 2y\) | |
| Norman Window: \(x = 2y\), \(r = \dfrac{x}{2}\), \(h = x\) | |
| Window with Isosceles Top: \(x = y\) | |
| Circular Sector: \(r = \sqrt{A}\), \(\theta = 2\ \text{rad}\) | |
| Right Triangle: \(x = y\), \(\theta = 45^\circ\) | |
| Closed Cylinder: \(D = h\) (or \(h = 2r\)) | |
| Open Cylinder: \(r = h\) (or \(D = 2h\)) | |
| Open Square Box: \(x = 2y\), \(x = \sqrt{\dfrac{A_{\text{surface}}}{3}}\) | |
| Closed Rectangular Box: \(x = y = z\) (cube) | |
| Stiffest Beam from Log: \(y = x\sqrt{3}\) | |
| Strongest Beam from Log: \(y = x\sqrt{2}\) | |
| Beam from Elliptic Section: \(x = \dfrac{2b}{\sqrt{3}}\), \(y = \dfrac{2\sqrt{2}a}{\sqrt{3}}\) | |
| Efficient Trapezoidal Section: \(x = 2y\), \(\theta = 120^\circ\) (half-hexagon) | |
| Cone in Sphere: \(h = \dfrac{4}{3}R\) | |
| Cylinder in Sphere: \(h = \dfrac{2R}{\sqrt{3}}\), \(V = \dfrac{4}{\sqrt{27}}\pi R^3\) | |
| Cylinder in Cone: \(y = \dfrac{h}{3}\) | |
| Cone of Given Slant Height: \(h = \dfrac{s}{\sqrt{3}}\), \(\theta = \tan^{-1}\sqrt{2}\) | |
| Max Viewing Angle: \(x = \sqrt{ab}\) | |
| Ladder Over Wall: \(L^{2/3} = h^{2/3} + a^{2/3}\) | |
| Beam Turning Perpendicular Corridor: \(L = (a^{2/3} + b^{2/3})^{3/2}\) | |
| Max Product \(x^m y^n\): \(x = \dfrac{m K}{m + n}\), \(y = \dfrac{n K}{m + n}\) | |
| Tank Cost Optimization: \(h = \left(\dfrac{\text{Cost}_{\text{ends}}}{\text{Cost}_{\text{sides}}}\right) D\) |
Worked Examples
Example. Find two positive numbers whose sum is \(60\) such that the product of one number and the square of the other is a maximum.
Solution. Let the numbers be \(x\) and \(y\), where \(x + y = 60 \implies y = 60 - x\).
The product function is \(P = x^2 y = x^2(60 - x) = 60x^2 - x^3\).
Differentiating with respect to \(x\) and equating to zero:
Factoring gives \(3x(40 - x) = 0\). Since \(x > 0\), \(x = 40\) and \(y = 60 - 40 = 20\).
Answer: \(40\ \text{and}\ 20\)
Example. A rectangular enclosure is constructed against a straight wall requiring no fencing along the wall. If \(160\ \text{m}\) of fencing is available, determine the dimensions for maximum area.
Solution. Let \(x\) be the length parallel to the wall and \(y\) be the width.
Perimeter constraint is \(x + 2y = 160 \implies x = 160 - 2y\).
Area function \(A = x y = (160 - 2y)y = 160y - 2y^2\).
Differentiating with respect to \(y\) and setting to zero:
Solving yields \(y = 40\ \text{m}\) and \(x = 160 - 2(40) = 80\ \text{m}\).
Answer: \(80\ \text{m} \times 40\ \text{m}\)
Example. An open box is made from a square sheet of side \(12\ \text{cm}\) by cutting equal squares of side \(x\) from each corner and turning up the sides. Find \(x\) for maximum volume.
Solution. Base dimensions are \(12 - 2x\) by \(12 - 2x\) with height \(x\).
Volume function \(V = (12 - 2x)^2 x = 144x - 48x^2 + 4x^3\).
Differentiating with respect to \(x\) and equating to zero:
Dividing by \(12\) yields \(x^2 - 8x + 12 = 0 \implies (x - 2)(x - 6) = 0\). Since \(x < 6\), \(x = 2\ \text{cm}\).
Answer: \(2\ \text{cm}\)
Example. A closed cylindrical can has a volume of \(128\pi\ \text{cm}^3\). Material for the top and bottom costs twice as much per unit area as the side. Find height \(h\) for minimum cost.
Solution. Volume \(V = \pi r^2 h = 128\pi \implies h = \dfrac{128}{r^2}\).
Cost function \(C = 2k(2\pi r^2) + k(2\pi r h) = 4\pi k r^2 + \dfrac{256\pi k}{r}\).
Differentiating with respect to \(r\) and setting to zero:
Solving gives \(8r^3 = 256 \implies r^3 = 32 \implies r = \sqrt[3]{32}\ \text{cm}\).
Height \(h = \dfrac{128}{(32)^{2/3}} = 8\sqrt[3]{4}\ \text{cm}\).
Answer: \(8\sqrt[3]{4}\ \text{cm}\)
Time Rates
Key Ideas
- Differentiate equations implicitly with respect to time
- Substitute instantaneous boundary values after differentiation
- Relate variables using geometric and physical formulas
Time Rates involve determining the instantaneous rate of change of a dependent variable with respect to time using differential calculus.
General Steps in Solving Time Rates
- Draw figure and label variable quantities
- Formulate equation connecting all variables
- Differentiate equation implicitly with respect to time
- Substitute given instantaneous values into derivative
Rate Definitions and Geometric Formulas
Linear Motion Rates
Volumetric Discharge Rate
Orthogonal Separation Rate
Oblique Angle Separation Rate
Conical Cistern Rate
Cylindrical Vessel Rate
Angular Elevation Rate
Shadow Lengthening Rate
Expanding Sphere Rate
Worked Examples
Example. Car A travels West at \(30\ \text{kph}\) starting at 12:00 PM. Car B starts from the same point at 2:00 PM and travels North at \(45\ \text{kph}\). Find how fast they are separating at 4:00 PM.
Solution. At 4:00 PM, time elapsed for Car A is \(t_A = 4\ \text{hr}\) and for Car B is \(t_B = 2\ \text{hr}\).
Distances traveled: \(x = 30(4) = 120\ \text{km}\) and \(y = 45(2) = 90\ \text{km}\).
Total separation \(s = \sqrt{x^2 + y^2} = \sqrt{120^2 + 90^2} = 150\ \text{km}\).
Differentiating \(s^2 = x^2 + y^2\) with respect to time:
Substituting values:
Answer: \(51\ \text{kph}\)
Example. Car A moves East at \(30\ \text{kph}\). At the same instant, Car B moves \(\text{S}\ 30^\circ\ \text{E}\) at \(60\ \text{kph}\). If the initial distance between them is zero, find how fast they are separating after \(1\ \text{hour}\).
Solution. The angle between East and \(\text{S}\ 30^\circ\ \text{E}\) is \(\theta = 60^\circ\).
After \(t = 1\ \text{hr}\), \(x = 30(1) = 30\ \text{km}\) and \(y = 60(1) = 60\ \text{km}\).
By Law of Cosines, \(s^2 = x^2 + y^2 - 2xy \cos 60^\circ = 30^2 + 60^2 - 2(30)(60)(0.5) = 2700 \implies s = 30\sqrt{3}\ \text{km}\).
Differentiating with respect to time \(t\):
Substituting values:
Answer: \(51.96\ \text{kph}\)
Example. Water flows into an inverted conical tank of height \(12\ \text{m}\) and top radius \(6\ \text{m}\) at a rate of \(8\ \text{m}^3/\text{min}\). How fast is the water level rising when the water is \(4\ \text{m}\) deep?
Solution. By similar triangles, \(\dfrac{r}{h} = \dfrac{6}{12} = \dfrac{1}{2} \implies r = \dfrac{h}{2}\).
Volume \(V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi \left(\dfrac{h}{2}\right)^2 h = \dfrac{\pi h^3}{12}\).
Differentiating with respect to time \(t\):
Substituting \(\dfrac{dV}{dt} = 8\) and \(h = 4\):
Answer: \(0.637\ \text{m/min}\)
Example. A balloon is released from the ground \(100\ \text{m}\) from an observer. If the balloon rises vertically at \(4\ \text{m/s}\), how fast is the balloon receding from the observer \(10\ \text{seconds}\) later?
Solution. Let \(y\) be height of balloon and \(s\) be distance from observer. \(s^2 = 100^2 + y^2\).
At \(t = 10\ \text{s}\), \(y = 4(10) = 40\ \text{m}\).
Distance \(s = \sqrt{100^2 + 40^2} = \sqrt{11600} \approx 107.7\ \text{m}\).
Differentiating \(s^2 = 10000 + y^2\) with respect to time:
Substituting \(y = 40\), \(s = 107.7\), and \(\dfrac{dy}{dt} = 4\):
Answer: \(1.49\ \text{m/s}\)
Example. A boy drinks cola from a cylindrical glass \(6\ \text{cm}\) in diameter at a rate of \(6\ \text{cm}^3/\text{min}\). How fast is the liquid level decreasing?
Solution. Radius \(r = 3\ \text{cm}\). Volume \(V = \pi r^2 h = 9\pi h\).
Differentiating with respect to time \(t\):
Given \(\dfrac{dV}{dt} = -6\ \text{cm}^3/\text{min}\):
Answer: \(0.212\ \text{cm/min}\)
Example. A man \(1.8\ \text{m}\) tall walks away from a street lamp \(6\ \text{m}\) high at a rate of \(1.2\ \text{m/s}\). How fast is his shadow lengthening?
Solution. By similar triangles, \(\dfrac{x + s}{6} = \dfrac{s}{1.8} \implies 1.8(x + s) = 6s \implies 1.8x = 4.2s \implies s = \dfrac{3}{7}x\).
Differentiating with respect to time \(t\):
Substituting \(\dfrac{dx}{dt} = 1.2\ \text{m/s}\):
Answer: \(0.514\ \text{m/s}\)
Formula Summary: Differential Calculus Applications
Key Ideas
- Summarizes all critical optimization and time rate equations
- Provides standard geometric proportions for maximum efficiency
- Serves as quick reference for engineering board exam problems
| Formula Name | Formula |
|---|---|
| Relative Maximum Condition | \(y' = 0\) and \(y'' < 0\) |
| Relative Minimum Condition | \(y' = 0\) and \(y'' > 0\) |
| Point of Inflection Condition | \(y'' = 0\) |
| Rectangle Inscribed in Circle | \(x = y = r\sqrt{2}\) |
| Rectangle Inscribed in Semicircle | \(b = 2h\), \(x = r\sqrt{2}\), \(y = \dfrac{r}{\sqrt{2}}\) |
| Rectangle Inscribed in Triangle | \(x = \dfrac{b}{2}\), \(y = \dfrac{h}{2}\) |
| Rectangle Inscribed in Ellipse | \(x = \dfrac{a}{\sqrt{2}}\), \(y = \dfrac{b}{\sqrt{2}}\) |
| Rectangle with Fixed Perimeter | \(x = y = \dfrac{P}{4}\) |
| Three-Sided Fenced Area | \(x = 2y\) |
| Norman Window Optimization | \(x = 2y\), \(r = \dfrac{x}{2}\), \(h = x\) |
| Window with Isosceles Top | \(x = y\) |
| Circular Sector Perimeter | \(r = \sqrt{A}\), \(\theta = 2\text{ rad}\) |
| Right Triangle Area Optimization | \(x = y\), \(\theta = 45^\circ\) |
| Closed Cylindrical Tank | \(D = h\) |
| Open Cylindrical Tank | \(r = h\) |
| Open Square-Based Box | \(x = 2y\), \(x = \sqrt{\dfrac{A}{3}}\) |
| Closed Rectangular Box | \(x = y = z\) |
| Stiffest Beam from Circular Log | \(y = x\sqrt{3}\) |
| Strongest Beam from Circular Log | \(y = x\sqrt{2}\) |
| Beam from Elliptic Section | \(x = \dfrac{2b}{\sqrt{3}}\), \(y = \dfrac{2\sqrt{2}a}{\sqrt{3}}\) |
| Most Efficient Trapezoidal Channel | \(x = 2y\), \(\theta = 120^\circ\) |
| Cone Inscribed in Sphere | \(h = \dfrac{4}{3}R\) |
| Cylinder Inscribed in Sphere | \(h = \dfrac{2R}{\sqrt{3}}\), \(V = \dfrac{4\pi R^3}{3\sqrt{3}}\) |
| Cylinder Inscribed in Cone | \(y = \dfrac{h}{3}\) |
| Cone of Maximum Volume | \(h = \dfrac{s}{\sqrt{3}}\), \(\theta = \tan^{-1}\sqrt{2}\) |
| Maximum Viewing Angle | \(x = \sqrt{ab}\) |
| Minimum Ladder Over Wall | \(L^{2/3} = h^{2/3} + a^{2/3}\) |
| Beam Turning Hallway Corner | \(L = (a^{2/3} + b^{2/3})^{3/2}\) |
| Product Optimization \(x^m y^n\) | \(x = \dfrac{m K}{m + n}\), \(y = \dfrac{n K}{m + n}\) |
| Instantaneous Velocity | \(v = \dfrac{ds}{dt}\) |
| Instantaneous Acceleration | \(a = \dfrac{dv}{dt} = \dfrac{d^2 s}{dt^2}\) |
| Volumetric Discharge Rate | \(Q = \dfrac{dV}{dt}\) |
| Perpendicular Separation Rate | \(s \dfrac{ds}{dt} = x \dfrac{dx}{dt} + y \dfrac{dy}{dt}\) |
| Oblique Separation Rate | \(s \dfrac{ds}{dt} = x \dfrac{dx}{dt} + y \dfrac{dy}{dt} - \left(x \dfrac{dy}{dt} + y \dfrac{dx}{dt}\right)\cos\theta\) |
| Conical Cistern Level Rate | \(\dfrac{dV}{dt} = \dfrac{\pi R^2}{H^2} h^2 \dfrac{dh}{dt}\) |
| Cylindrical Vessel Level Rate | \(\dfrac{dV}{dt} = \pi r^2 \dfrac{dh}{dt}\) |
| Angle of Elevation Rate | \(\dfrac{dy}{dt} = x \sec^2\theta \dfrac{d\theta}{dt}\) |
| Shadow Growth Rate | \(\dfrac{ds}{dt} = \dfrac{h}{H - h} \dfrac{dx}{dt}\) |
| Spherical Volume Expansion Rate | \(\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}\) |
| Spherical Surface Expansion Rate | \(\dfrac{dA}{dt} = 8\pi r \dfrac{dr}{dt}\) |