Ｃａｌｃｕｌｕｓ　１　ｌｅｃｃｔｕｒｅ　３．７

　Oｐｔｉｍｉｚａｔｉｏｎ；　Mａｘ / Mｉｎ　Aｐｐｌｉｃａｔｉｏｎ　Pｒｏｂｌｅｍｓ

Ｉｎｔｒｏｄｕｃｔｉｏｎ

• [0:01]. Introduction - [📷image]
  
  • This lecture focuses on how to maximize and/or minimize continuous functions.
  • Two scenarios are explored:
    • Closed Intervals.
    • Open Intervals.
  • We analyze the existence of absolute maxima and minima in continuous functions and their real-life applications.
• [0:50]. Real-Life Applications
  • Minimizing Costs and Maximizing Profits: In business, it is crucial to find the lowest cost and the highest profit.
    • Calculus allows for precise calculations of production and costs to optimize these aspects.
  • Design Optimization: Calculus can be used to design products that maximize attributes such as volume, given certain constraints.

Ｅｘｔｒｅｍｅ　ｖａｌｕｅ　ｔｈｅｏｒｅｍ

• [2:15]. Extreme Value Theorem.
  • Closed Intervals:
    • If a function is continuous on a closed interval $[\mathcal{a},\mathcal{b}]$, then it must attain both an absolute maximum and an absolute minimum on that interval.
    • These extrema may occur:
      • At the endpoints ($x=\mathcal{a}$ or $x=\mathcal{b}$), or
      • At critical points within $(\mathcal{a}, \mathcal{b})$ where the derivative is zero or undefined.
  • Open Intervals:
    • If a function is continuous on an open interval $(\mathcal{a},\mathcal{b})$, it might not attain absolute extrema.
    • This is because there are no endpoints to "trap" the function’s values, and the function might increase or decrease without bound near the edges.

Ｍａｘｉｍｉｚｉｎｇ　Eｘａｍｐｌｅｓ

• [3:51]. Maximizing the Area of a Rectangle - [📷image]
  • Problem: Maximize the area of a rectangular region fenced with 100 feet of fencing.
  • Identify the Restriction: The total length of fencing (100 ft) corresponds to the rectangle’s perimeter.
  • Draw a Diagram: Sketch a rectangle and label its sides as $𝒙$ and $𝑦$.
  • Formulate:
    • Area: $A = 𝒙\cdot 𝑦$
  • Constraint:
    • Perimeter: $2𝒙 + 2𝑦 = 100$
      • Solve the restriction for one variable:
      • $𝑦 = 50 − 𝒙$
      • Substitute into the Area Formula: Substitute $𝑦$ into the area expression:
        • $A = 𝒙(50 − 𝒙)$
  • Find Critical Points: Take the derivative of $𝒜$ with respect to $𝒙$, set it to zero, and solve for $𝒙$
    • $ \dfrac{dA}{d𝒙} = 50 − 2𝒙 = 0$
    • $𝒙 = 25$
  • Determine Endpoints: The possible minimum and maximum values for $𝒙$ are $0$ and $50$.
    • [18:08]. ᴛʜᴇ ᴩᴏɪɴᴛ: By the Extreme Value Theorem, the maximum must occur at one of the following:
      • $𝒙 = 0,\; 𝒙 = 25,\; 𝒙 = 50$ — all within the interval $𝒙 \in [0, 50]$
  • Evaluate the Function at Critical Points and Endpoints:
    • $A(0) = 0$
    • $A(25) = 625$
    • $A(50) = 0$
  • Conclusion: The maximum area is $625\ \text{ft}^2$ when $𝒙 = 25$ ft and $𝑦 = 25$ ft, forming a square.
• [21:41]. Maximizing the Volume of a Box - [📷image-1] - [📷image-2]
  • Problem: An open box is constructed from a $16\times 30$-inch piece of cardboard by cutting squares of side $𝒙$ from each corner and folding up the flaps.
  • Draw a Diagram: Sketch the cardboard with the cut-out squares and the resulting box.
  • Identify the Restriction: The maximum cut size is $8$ inches (half the length of the shorter side).
  • Formulate Equations:
    • Volume: $V = 𝒙(30 − 2𝒙)(16 − 2𝒙)$
    • Restriction: $0 \le 𝒙 \le 8$
  • Simplify the Volume Formula: Expand and combine like terms as needed.
  • Find Critical Points: Take the derivative of $V$ with respect to $𝒙$, set it to zero, and solve.
    • $ \dfrac{dV}{d𝒙} = 12𝒙^2 − 188𝒙 + 480 = 0$
    • By factoring or quadratic formula: $𝒙 = 12$ or $𝒙 = \dfrac{10}{3}$
  • Discard Invalid Solutions: $𝒙 = 12$ is invalid because it exceeds the restriction.
  • Evaluate the Function at Critical Points and Endpoints:
    • $V(0) = 0$
    • $V\!\left(\dfrac{10}{3}\right) > 0$
    • $V(8) = 0$
  • Conclusion: The maximum volume is achieved with a $10/3$-inch cut.

Ｍｉｎｉｍｉｚｉｎｇ　Eｘａｍｐｌｅｓ

• [38:33]. Minimizing the Cost of a Pipeline - [📷image-1] - [📷image-2]
  • Problem: A pipeline must carry oil from an offshore platform to a refinery on land.
    • The pipeline cost is $1 per km offshore and $0.50 per km on land.
    • The platform is $5$ km from the coast, and the refinery is $8$ km along the coast.
  • Draw a Diagram: Sketch the coastline, the offshore platform, the refinery, and the potential pipeline route.
  • Identify the Restriction: The refinery is fixed at $8$ km along the coast.
  • Choose a landing point $𝓟=(x,0)$ somewhere along the coast ($0 \le x \le 8$) where the total cost is minimized.
  • Formulate Equations:
    • Cost: $C = \sqrt{x^{2} + 25}\; +\; 0.5\,(8 − x)$
      (Here, $x$ is the distance along the coast from the point directly opposite the platform to the pipeline’s landfall)
    • Restriction: $0 \le x \le 8$
  • Simplify the Cost Equation: Distribute $0.5$ in the expression.
  • Find Critical Points: Take the derivative of $C$ with respect to $x$, set it to zero, and solve.
    • $ \dfrac{dC}{dx} = \dfrac{x}{\sqrt{x^{2} + 25}} − 0.5 = 0$
    • Solving yields $x = \dfrac{5\sqrt{3}}{3}$
  • Evaluate the Function at Critical Points and Endpoints:
    • $C(0) = 9$
    • $C\!\left(\dfrac{5\sqrt{3}}{3}\right) \approx 8.33$
    • $C(8) \approx 9.43$
  • Conclusion: The minimum cost is about 8.33$ when x $\approx$ 2.9 km.
• [1:12:12]. Minimizing Material for a Can - [📷image-1] - [📷image-2] - [📷image-3]
  • Problem: Design a cylindrical can that holds $1{,}000\ \text{cm}^3$ of liquid using the least material.
  • Draw a Diagram: Sketch a cylinder, labeling the radius $r$ and height $h$.
  • Identify the Restriction: The can’s volume is fixed at $1{,}000\ \text{cm}^3$.
  • Formulate Equations:
    • Surface Area: $S = 2\pi r^{2} + 2\pi r\,h$
    • Restriction (Volume): $\pi r^{2}h = 1000$
  • Solve for One Variable: Solve the volume restriction for $h$.
    • $h = \dfrac{1000}{\pi r^{2}}$
  • Substitute into the Surface Area Formula:
    • $S = 2\pi r^{2} + 2\pi r \left(\dfrac{1000}{\pi r^{2}}\right)$
    • Simplify: $S = 2\pi r^{2} + \dfrac{2000}{r}$
  • Find Critical Points: Take the derivative of $S$ with respect to $r$, set it to zero, and solve.
    • $ \dfrac{dS}{dr} = 4\pi r − \dfrac{2000}{r^{2}} = 0$
    • Solving gives $r = \sqrt[3]{\dfrac{500}{\pi}} \approx 5.42\ \text{cm}$
  • Calculate the Height:
    • $h \approx 10.84\ \text{cm}$
  • Verify It’s a Minimum:
    • Use the second derivative: $\dfrac{d^{2}S}{dr^{2}} = 4\pi + \dfrac{4000}{r^{3}}$
    • Evaluating at $r \approx 5.42$ cm yields a positive result, indicating a minimum.
  • Conclusion: The can uses the least material when $r \approx 5.42$ cm and $h \approx 10.84$ cm.