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Muskan

  • 2 years ago

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

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  1. helder_edwin
    • 2 years ago
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    creo que parafraseaste mal el problema

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

  3. Muskan
    • 2 years ago
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    is this a answer... but i dont understand

  4. helder_edwin
    • 2 years ago
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    qué no entiendes?

  5. jasonxx
    • 2 years ago
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    thats all ya need to do, ya need to differentiate

  6. Muskan
    • 2 years ago
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    la letra...

  7. jasonxx
    • 2 years ago
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    what?

  8. helder_edwin
    • 2 years ago
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    :). sí. es un poco difícil de leer.

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

  10. Muskan
    • 2 years ago
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    no me gusta la optimización...no me entero nada...

  11. helder_edwin
    • 2 years ago
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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.

  12. helder_edwin
    • 2 years ago
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    sólo tienes que aplicar derivadas, para optimizar.

  13. helder_edwin
    • 2 years ago
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    entiendes hasta aquí?

  14. Muskan
    • 2 years ago
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    hmmm

  15. helder_edwin
    • 2 years ago
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    sí o no?

  16. Muskan
    • 2 years ago
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    mas o menos

  17. helder_edwin
    • 2 years ago
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    esto lo aprendiste en geometría, en el colegio.

  18. Muskan
    • 2 years ago
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    yeah ...bueno intentare....

  19. helder_edwin
    • 2 years ago
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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 \[ \large 2x+2y=8 \]

  20. Muskan
    • 2 years ago
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    vale

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

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

  23. helder_edwin
    • 2 years ago
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    muy bien. cuánto es la base en este triángulo?

  24. Muskan
    • 2 years ago
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    2x

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

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

  27. helder_edwin
    • 2 years ago
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    simplifica.

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

  29. helder_edwin
    • 2 years ago
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    simplifica

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

  31. helder_edwin
    • 2 years ago
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    @Muskan ¿entendiste?

  32. Muskan
    • 2 years ago
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    como x * h

  33. Muskan
    • 2 years ago
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    ahh vale

  34. Muskan
    • 2 years ago
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    ya lo pio

  35. helder_edwin
    • 2 years ago
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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.

  36. Muskan
    • 2 years ago
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    hmmm

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

  38. Muskan
    • 2 years ago
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    no se

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

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

  41. helder_edwin
    • 2 years ago
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    |dw:1355064338201:dw|

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

  43. helder_edwin
    • 2 years ago
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    muy bien. despeja h.

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

  45. helder_edwin
    • 2 years ago
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    muy bien. reemplaza esto en la fórmula del área.

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

  47. helder_edwin
    • 2 years ago
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    muy bien. recuerdas la ecuación del perímetro?

  48. Muskan
    • 2 years ago
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    si

  49. Muskan
    • 2 years ago
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    x + y = 4

  50. helder_edwin
    • 2 years ago
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    despeja "y" y reemplaza en la ecuación del área.

  51. Muskan
    • 2 years ago
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    y = x + 4

  52. helder_edwin
    • 2 years ago
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    no

  53. Muskan
    • 2 years ago
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    x - 4

  54. helder_edwin
    • 2 years ago
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    y=x-4. reemplaza

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

  56. helder_edwin
    • 2 years ago
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    muy mal

  57. helder_edwin
    • 2 years ago
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    inténtalo de nuevo.

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

  59. helder_edwin
    • 2 years ago
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    te falta algo.

  60. helder_edwin
    • 2 years ago
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    la potencia de "y"

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

  62. Muskan
    • 2 years ago
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    es muy largo...

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

  64. Muskan
    • 2 years ago
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    hmm..bueno intentare sola..

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