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Muskan

calculates the base and the height of the isosceles triangle of perimeter and area 8 maximum.

  • one year ago
  • one year ago

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  1. helder_edwin
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    creo que parafraseaste mal el problema

    • one year ago
  2. jasonxx
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    |dw:1355062158969:dw|\[x*y=8\] y=8/x now i guess ya can do it

    • one year ago
  3. Muskan
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    is this a answer... but i dont understand

    • one year ago
  4. helder_edwin
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    qué no entiendes?

    • one year ago
  5. jasonxx
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    thats all ya need to do, ya need to differentiate

    • one year ago
  6. Muskan
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    la letra...

    • one year ago
  7. jasonxx
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    what?

    • one year ago
  8. helder_edwin
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    :). sí. es un poco difícil de leer.

    • one year ago
  9. helder_edwin
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    un triángulo isóceles tiene dos lados iguales. el tercero hará de base.

    • one year ago
  10. Muskan
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    no me gusta la optimización...no me entero nada...

    • one year ago
  11. helder_edwin
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    la longitud de los lados iguales será "y" como en la figura del papel. si trazas la altura desde el vértice que forman los lados iguales de un triángulo isóceles hacia el otro lado (que hace de base) divides el triángulo en dos triángulos rectángulos.

    • one year ago
  12. helder_edwin
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    sólo tienes que aplicar derivadas, para optimizar.

    • one year ago
  13. helder_edwin
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    entiendes hasta aquí?

    • one year ago
  14. Muskan
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    hmmm

    • one year ago
  15. helder_edwin
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    sí o no?

    • one year ago
  16. Muskan
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    mas o menos

    • one year ago
  17. helder_edwin
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    esto lo aprendiste en geometría, en el colegio.

    • one year ago
  18. Muskan
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    yeah ...bueno intentare....

    • one year ago
  19. helder_edwin
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    ahora el perímetro tiene que ser 8. pero el perímetro es la suya de las longitudes de todos los lados. entonces tienes la primera ecuación \[ \large 2x+2y=8 \]

    • one year ago
  20. Muskan
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    vale

    • one year ago
  21. helder_edwin
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    tu objetivo es maximizar el área del triángulo isóceles. cuál es el área del triángulo?

    • one year ago
  22. Muskan
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    a= 1/2 * b * h

    • one year ago
  23. helder_edwin
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    muy bien. cuánto es la base en este triángulo?

    • one year ago
  24. Muskan
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    2x

    • one year ago
  25. helder_edwin
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    muy bien. ahore escribe de nuevo tu fórmula del área reemplazando b=2x

    • one year ago
  26. Muskan
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    a= 1/2 * 2x * h asi?

    • one year ago
  27. helder_edwin
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    simplifica.

    • one year ago
  28. Muskan
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    a = 2x/2 * h

    • one year ago
  29. helder_edwin
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    simplifica

    • one year ago
  30. helder_edwin
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    \[ \large A=1/2\cdot2x\cdot h=x\cdot h \]

    • one year ago
  31. helder_edwin
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    @Muskan ¿entendiste?

    • one year ago
  32. Muskan
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    como x * h

    • one year ago
  33. Muskan
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    ahh vale

    • one year ago
  34. Muskan
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    ya lo pio

    • one year ago
  35. helder_edwin
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    ahora. siempre recuerda que el problema tienes que resolverlo con las herramientas que tienes. como estás haciendo cálculo en una variable, las técnicas que conoces sólo las puedes aplicar cuando tienes una sola variable.

    • one year ago
  36. Muskan
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    hmmm

    • one year ago
  37. helder_edwin
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    en la fórmula \[ \large A=x\cdot h \] tienes dos variables. qué haces?

    • one year ago
  38. Muskan
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    no se

    • one year ago
  39. helder_edwin
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    te deshaces de una de ellas. la pregunta ahora es ¿cómo?

    • one year ago
  40. helder_edwin
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    recuerdas que el triángulo isocéles quedó divido en dos triángulos rectángulos idénticos. y que están dibujados en la hoja.

    • one year ago
  41. helder_edwin
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    |dw:1355064338201:dw|

    • one year ago
  42. Muskan
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    \[y ^{2} = x ^{2} + h ^{2}\]

    • one year ago
  43. helder_edwin
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    muy bien. despeja h.

    • one year ago
  44. Muskan
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    h = \[h = \sqrt{y ^{2} - x ^{2}}\]

    • one year ago
  45. helder_edwin
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    muy bien. reemplaza esto en la fórmula del área.

    • one year ago
  46. Muskan
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    \[a = x *\sqrt{ y ^{2} - x ^{2}}\]

    • one year ago
  47. helder_edwin
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    muy bien. recuerdas la ecuación del perímetro?

    • one year ago
  48. Muskan
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    si

    • one year ago
  49. Muskan
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    x + y = 4

    • one year ago
  50. helder_edwin
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    despeja "y" y reemplaza en la ecuación del área.

    • one year ago
  51. Muskan
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    y = x + 4

    • one year ago
  52. helder_edwin
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    no

    • one year ago
  53. Muskan
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    x - 4

    • one year ago
  54. helder_edwin
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    y=x-4. reemplaza

    • one year ago
  55. Muskan
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    \[a = x * \sqrt{x-4} - x ^{2}\]

    • one year ago
  56. helder_edwin
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    muy mal

    • one year ago
  57. helder_edwin
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    inténtalo de nuevo.

    • one year ago
  58. Muskan
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    \[a = x * \sqrt{(x-4) - x ^{2}}\]

    • one year ago
  59. helder_edwin
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    te falta algo.

    • one year ago
  60. helder_edwin
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    la potencia de "y"

    • one year ago
  61. Muskan
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    \[a = x * \sqrt{(x-4)^{2} - x ^{2}}\]

    • one year ago
  62. Muskan
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    es muy largo...

    • one year ago
  63. helder_edwin
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    muy bien. viste. ahora tienes una función con una solo variable. y apuedes aplicar cálculo.

    • one year ago
  64. Muskan
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    hmm..bueno intentare sola..

    • one year ago
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