[rayyrayy]

Ok so I have failed to complete any of the word problems given to me in my Algebra II book, because I can not figure out how to manipulate the equations to get the correct information.

First problem I sort of figured but I'm still having trouble.

1. Find two numbers whose product is 120 and whose sum is a minimum.

I got - -

(x) * (120/x) = 120 so Sum(x) = x + 120/x which is right but how do I figure out whether it is a minimum or a maximum?? Do I put it into the formula y = a(x - h)^2 + k?? And if so how?


I'm trying to use the information from this problem to solve the next one which is NOT going well each time I come up against difficulties.

2. Find two numbers who difference is 8 and who product is a minimum. (let x = the first number and x + 8 = the second number)

Help me please I've spent 3 useless days on this problem, and now my life seems meaningless.

[red_cartoon]

Do you have any constraints like your results must be integers or something like that ?

Are you allowed to use calculus ? If you are, then differentiate the expressions you have figured out , look for a minima. Hope it helps.

[terryg752]

1. Let numbers be x, 120/x

Sum = y = x + 120/x

For minimum, dy/dx = 0, d^2y/dx^2 is positive.

dy/dx = 1 - 120/x^2

= 0 if x^2 = 120

x = sqrt (120)

d^2y/dx^2 = 240/x^3

At x = sqrt(120), d^2y/dx^2 is positive.

Hence x = sqrt (120) gives a minimum.

The other number = 120/x = sqrt (120)

Numbers are sqrt (120), sqrt (120)


Do similar procedure for second problem.

[KISS]

I think you guys blew Ray away. Algebra II comes before Calculus. You might be scareing the poor person.

It does sound like a Calculus problem to me too.

[red_cartoon]

really, I am sorry then. I guessed so, that's why asked whether you are allowed to use calculus or not. Anyway, here is a non-calculus solution for you second question.

say one of the numbers is x and the other is x+8. Then their product p will be,

p=x*(x+8)
= x^2 + 8x
= x^2 + 2*x*4 + 4^2 - 4^2
= ( x + 4 )^2 - 4^2

-4^2 here is a constant. So we have to rely on ( x + 4 )^2 for the minimization task at hand. Since this is a squared term, the minimum value it can obtain is 0. From this reasoning we can say,

( x + 4 )^2 = 0
or, x + 4 = 0
or, x = -4

using this result you can find the other number, x+8 = 4.

So the numbers you were looking for are 4 and -4.

Hope this helps you :)

[KISS]

Red:

The question stated "in my Algebra II book". Implicitly, that means no Calculus.

[red_cartoon]

Yeah, you are right. I should have understood that. But since I am not familiar with 'Algebra II' book, I decided to take a chance. In what class/standard/stage do they teach Algebra II ? High school ?

[pskinner]

Ok so I have failed to complete any of the word problems given to me in my Algebra II book, because I can not figure out how to manipulate the equations to get the correct information.

First problem I sort of figured but I'm still having trouble.

1. Find two numbers whose product is 120 and whose sum is a minimum.

I got - -

(x) * (120/x) = 120 so Sum(x) = x + 120/x which is right but how do I figure out whether it is a minimum or a maximum??? Do I put it into the formula y = a(x - h)^2 + k ??? and if so how?


I'm trying to use the information from this problem to solve the next one which is NOT going well each time I come up against difficulties.

2. Find two numbers who difference is 8 and who product is a minimum. (let x = the first number and x + 8 = the second number)

Help me please I've spent 3 useless days on this problem, and now my life seems meaningless.

The numbers are x and 120-x, so those are the number to put into:

x*(120-x)=120.

For a good explanation of how to solve this or any quadratic equation, go to:
Free Online Quadratic Equation Solver: Solve by Most Efficient Method

[KISS]

A long, long, long time ago, I remember algebra, algebra I, algebra II, Geometry, Pre-calc, Calculus, differential equations. With logic, statistics and probability thrown in someplace.

There is overlap, but it definitely progressed in that order.