# Parabola min/max word problem using vertex form

• Sep 30th 2007, 10:29 AM
ruscutie100
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. :confused:

Please explain it because i don't know how to do these kind of questions... I have a test tomorrow...plz help.. :)
• Sep 30th 2007, 10:33 AM
Jhevon
Quote:

Originally Posted by ruscutie100
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. :confused:

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?
• Sep 30th 2007, 10:39 AM
ruscutie100
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
• Sep 30th 2007, 10:46 AM
Jhevon
Quote:

Originally Posted by ruscutie100
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 $x = 24 - 2y$

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

$\Rightarrow \boxed {x = 12}$

now, can you state the maximum product?
• Sep 30th 2007, 11:02 AM
ruscutie100
Quote:

Originally Posted by Jhevon
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 $x = 24 - 2y$

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

$\Rightarrow \boxed {x = 12}$

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

= -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"

• Sep 30th 2007, 11:46 AM
Jhevon
Quote:

Originally Posted by ruscutie100
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