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? 發問於 Science & MathematicsMathematics · 5 年前

Calc I Min/Max Problem?

Can someone explain to me how to set this problem up? And what derivative am I looking for?

Find two positive numbers that satisfy the given requirements.

The sum of the first number squared and the second number is 60 and the product is a maximum.

4 個解答

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  • 最愛解答

    You have 2 numbers, x and y

    x^2 + y = 60

    y = 60 - x^2

    P = x * y

    P = x * (60 - x^2)

    P = 60x - x^3

    Derive P with respect to x

    dP/dx = 60 - 3x^2

    dP/dx = 0

    0 = 60 - 3x^2

    3x^2 = 60

    x^2 = 20

    x = +/- 2 * sqrt(5)

    x > 0, so x = 2 * sqrt(5)

    x^2 + y = 60

    20 + y = 60

    y = 40

  • cidyah
    Lv 7
    5 年前

    Let x and y be the two numbers

    x^2 + y = 60

    Product = xy is maximum

    Product = xy

    Express y in terms of x

    Product = x (60-x^2)

    Product = 60x -x^3

    dP/dx = 60 -3x^2 = 0

    3x^2 = 60

    x^2 = 20

    x = ± √20

    d^2P/dx^2 = -6x

    when x= √20, d^2P/dx^2 = -6 (√20) < 0 , so Product has a maximum at x= √20

    The numbers are x = √20

    y= 60 - x^2 = 60- 20 = 40

    The numbers are √20 and 40

    The maximum product is : (40)(√20) = 178.8854

  • Raj K
    Lv 7
    5 年前

    Let the numbers be x and y

    →x²+y=60

    i.e. y=60−x²

    Product P=xy

    P=xy=x(60−x²) =60x−x³

    dP/dx= 60−3x² =0 for maxima or minima

    →x²=20and x=√20=2√5 Since numbers are positive

    Hence y=(60−x²)=(60−20)=40

    d²P/dx²= −6x negative for positive x Hence the Value of x is for maximum P

    Maximum product P=xy=2√5×40=80√5

  • ?
    Lv 7
    5 年前

    x > 0, y > 0

    x² + y = 60

    y = 60 - x²

    find max P = xy

    = x(60 - x²) = 60x - x³

    P' = dP/dx = 60 - 3x² = 0

    3x² = 60

    x² = 20

    x = ±√20, reject negative value

    x = √20

    y = 60 - x² = 60 - 20 = 40

    y = 40

    P = xy = 40√20

    x = √20, y = 40

    P" = - 6x

    -6√20 < 0

    Therefore P is a maximum.

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