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Math Help - Parabola min/max word problem using vertex form

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    Post Parabola min/max word problem using vertex form

    The sum of a number and two times another number is 24. Find the numbers if their product is a maxiumum and state the maximum.

    Please explain it because i don't know how to do these kind of questions... I have a test tomorrow...plz help..
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ruscutie100 View Post
    The sum of a number and two times another number is 24. Find the numbers if their product is a maxiumum and state the maximum.

    Please explain it because i don't know how to do these kind of questions... I have a test tomorrow...plz help..
    Start with the basics. what are our unknowns here? the two numbers of course. so, they're unknowns, call them something.

    Let one number be x
    Let the other number be y

    ok, what are we told about x and y?

    the sum of x and twice y is 24. "sum" means "add". so we have:

    x + 2y = 24 .......................(1)

    ok, what now? We want their product to be a maximum. call the product P. product means to multiply, so we want:

    P = xy .............(2) to be a maximum

    so we have our constraint, which is equation (1), and we have the function we want to maximize, which is P

    now, we can take two routes to complete the question. which route we take depends on whether you can use calculus or not. so, can you?
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    I can't use calculus... my teacher setup the equation and then completed the square and found the maximum.... but i don't get it..


    thnx
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ruscutie100 View Post
    I can't use calculus... my teacher setup the equation and the completed the square and found the maximum.... but i don't get it..


    thnx
    ok, so we want the precalc way of doing this. that's fine.

    so we have x + 2y = 24 ..........(1)
    and P = xy ................(2)

    we want to maximize P. to do this, it would be much simpler to work with one variable. so solve for one variable in terms of the other using our constraint.

    From equation (1), x = 24 - 2y, so plug that into P

    \Rightarrow P = (24 - 2y)y = 24y - 2y^2

    this is a downward opening parabola, it's maximum value occurs at the vertex. we can find the vertex by completing the square, but that is too much work. use the vertex formula:

    Vertex formula: for a parabola y = ax^2 + bx + c, the x value for the vertex is given by: x = \frac {-b}{2a}, and thus the vertex is given by \left( \frac {-b}{2a}, f\left( \frac {-b}{2a} \right) \right)

    so here, the vertex occurs when y = \frac {-24}{2(-2)} = 6

    so \boxed {y = 6}

    but we have x = 24 - 2y

    \Rightarrow x = 24 - 2(6) = 12

    \Rightarrow \boxed {x = 12}

    there are your two numbers

    now, can you state the maximum product?
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    Quote Originally Posted by Jhevon View Post
    ok, so we want the precalc way of doing this. that's fine.

    so we have x + 2y = 24 ..........(1)
    and P = xy ................(2)

    we want to maximize P. to do this, it would be much simpler to work with one variable. so solve for one variable in terms of the other using our constraint.

    From equation (1), x = 24 - 2y, so plug that into P

    \Rightarrow P = (24 - 2y)y = 24y - 2y^2

    this is a downward opening parabola, it's maximum value occurs at the vertex. we can find the vertex by completing the square, but that is too much work. use the vertex formula:

    Vertex formula: for a parabola y = ax^2 + bx + c, the x value for the vertex is given by: x = \frac {-b}{2a}, and thus the vertex is given by \left( \frac {-b}{2a}, f\left( \frac {-b}{2a} \right) \right)

    so here, the vertex occurs when y = \frac {-24}{2(-2)} = 6

    so \boxed {y = 6}

    but we have x = 24 - 2y

    \Rightarrow x = 24 - 2(6) = 12

    \Rightarrow \boxed {x = 12}

    there are your two numbers

    now, can you state the maximum product?
    thnx... the max value is 72... i'm really new to this method (srry i don't get it) ... i used the equations u made... after that i completed the square... and got it in the form y = a(x-h)^2 + K

    so my answer was
    = -2 (y^2 -6)^2 + 72

    and 72 was the max... my teacher told me that 72 is the max when x-coordinate is -6... so the other number will be "-12"

    so is my answer right?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ruscutie100 View Post
    thnx... the max value is 72... i'm really new to this method (srry i don't get it) ... i used the equations u made... after that i completed the square... and got it in the form y = a(x-h)^2 + K

    so my answer was
    = -2 (y^2 -6)^2 + 72

    and 72 was the max... my teacher told me that 72 is the max when x-coordinate is -6... so the other number will be "-12"

    so is my answer right?
    y = +6 works as well. so it depends on what numbers you picked. if you choose y = -6, you have a different value for x than the one i told you, work it out.
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