APLICACION DE LA DERIVADA ONCE M2 S24

Summary

The lesson explains how derivatives can be used to analyze change and solve practical optimization problems. The first derivative helps locate critical points and describe where a function is increasing or decreasing. The second derivative can help classify a critical point as a local maximum or minimum.

Main concepts

• Applications of derivatives: The lesson gives examples from physics (speed and acceleration), economics (changes in cost, demand, and supply), engineering (control-system design), and biology (population growth).
• First derivative: Describes a function’s slope. Critical points can be found by solving (f’(x)=0), though these points must then be analyzed to determine their type.
• Second derivative test: At a critical point (c):
  • If (f’‘(c)>0), the graph is concave up there and has a local minimum.
  • If (f’‘(c)<0), the graph is concave down there and has a local maximum.
  • If (f’‘(c)=0), the test is inconclusive; another test, such as checking the sign of the first derivative, is needed.
• Optimization method: Build a function for the quantity to be maximized or minimized, differentiate it, solve (f’(x)=0), check the candidate with the second derivative or another appropriate test, and evaluate the original function to obtain the optimized value.

Worked applications

1. Profit from manufacturing

   • The function used is (f(x)=-x^2+100x+20).
   • Differentiating gives (f’(x)=-2x+100). Setting it to zero yields (x=50).
   • Substituting 50 into the original function gives a maximum profit of $2,520.
   • The subtitle’s final spoken figure of $520 conflicts with its own calculation: (-2500+5000+20=2520).

2. Apple production

   • Let (x) be the number of trees planted beyond the initial 50. The model is: [ P(x)=(50+x)(60-x)=3000+10x-x^2. ]

   • Solving (P’(x)=10-2x=0) gives (x=5). The second derivative is (-2), confirming a maximum.

   • The best planting is 55 trees per hectare, producing 3,025 kg of apples under the model.

3. Rectangular enclosure for fruit trees

   • With a perimeter of 400 m, the sides satisfy (2x+2y=400), so (y=200-x).
   • The area is (A(x)=x(200-x)=200x-x^2). Its maximum occurs at (x=100), and then (y=100).
   • The maximum area is 10,000 m². At 10 m² per tree, the enclosure can hold 1,000 trees.

Important correction to the subtitles

In the polynomial example, the function is (y=x^4+8x^3+22x^2+24x+9). Its derivative is (4(x+1)(x+2)(x+3)), giving critical points at (x=-1,-2,-3). The second-derivative test classifies (x=-1) and (x=-3) as local minima and (x=-2) as a local maximum. These are turning points, not inflection points, despite the wording in the subtitles.

The lesson closes by reviewing the derivative-based optimization process and inviting students to use the school’s tutoring and other academic support resources.

Speakers and sources

• Fabio Jiménez — presenter and instructor; parts of the introductory self-description are garbled in the auto-generated subtitles.
• “Prelude to Spring” — a poem excerpt read during the active break; the subtitles do not identify its author.
• Grupo Educativo Monterrosales — educational channel and source of the lesson.