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MarcLeclair

  • 3 years ago

- Optimization problem - A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola . What are the dimensions of such a rectangle with the greatest possible area?

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  1. MarcLeclair
    • 3 years ago
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    oh sorry the parabola is y=5-x^2

  2. MarcLeclair
    • 3 years ago
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    I was thinking find the 0 for the parabola, and because it touches the 2 upper part I get sqrt(5) and -sqrt(5) being the length for y and I can ind the length in x because those 2 y are the distance between 2 cornor that is the length?

  3. Chlorophyll
    • 3 years ago
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    Since the rectangular inscribed in the parabola: it's length is symmetry x+ x => A = 2x y = 2x ( 5 - x²)

  4. MarcLeclair
    • 3 years ago
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    oh. inscribed in, so i had it wrong... so you don't have any constraint in this problem...?

  5. Chlorophyll
    • 3 years ago
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    I think the range of parabola is the constraint, just get the concept first!

  6. MarcLeclair
    • 3 years ago
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    Well I usually go about these problem by finding my constraint then the equation I need to solve, but I couldn't do so. But when you say A=x(5-x^2) and so I take the derivative of both side and I should get x(5-2x) + (5-x^2) = A'(x) 5x-2x^2+5-x^2 -3x^2+5x+5 and then use quadratic to find my x ?

  7. Chlorophyll
    • 3 years ago
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    A = 2x ( 5 - x²) = -2x³ + 10x ->A' = -6x² + 10 = 0 => x = √ 5/3

  8. MarcLeclair
    • 3 years ago
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    but wait, why did you mention x + x being symmetrical? and I asked about the constraint because, don't you need one to do an optimization problem?

  9. Chlorophyll
    • 3 years ago
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    Did you sketch the rectangular inscribes in the parabola, you'll see the constraint -√ 5 ≤ x ≤ √5, 0 ≤ y ≤ 5

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