Find two positive real numbers whose difference is 6 and the product is a minimum?

Question

Use the Quadratic function... Show solutions please...

Answer 1

Four answers so far, and none of them are right.

You've said two positive real numbers, so that rules out any solution with -3 or zero. 1 and 7 could be right, but their product is 7. If you take 0.1 and 6.1 their product is 0.61, far less than 7. So 1 and 7 is not right.

I suspect maybe you have asked the question wrong, and that either you are looking for two integers, or two numbers only one of which is positive?

Otherwise the answer has as a limit 6.000000... and 0.0000000....
.

Answer 2

Let me try to solve this... and perhaps even prove if whether or not it is solvable.

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Lets just use any and all integers, for arguments sake... positive or negative.

Let s be the smaller number
Let l be the larger number.

their difference is six:
l - s = 6
Therefore,
l = s + 6

Their product is a minimum. Hmmm... well, lets start by just asking, what is their product?
P = s•l

If you combine the relationships that we know...
P = s•(s+6)

Little bit of algebra, rearranging...
P = s² + 6•s

Now, here is where it gets tricky... Im using a little calculus here. I dont expect you to follow. But remember, Im trying to figure out if there -is- a solution.

If I want to find the minimum of this parabola, P = s² + 6•s, then I have to find the point where is slope changes direction from increasing or decreasing to the opposite, and that is the point of its vertex.

I can do this with calculus... by deriving the function.
P' = 2•s + 6

At the minimum, P' will have a value of 0

So,
2•s + 6 = P'
2•s + 6 = 0
2•s = -6
s = -3

I conclude, then, that the smaller number is -3 while the larger is then +3.

This means their product will be -9... a given minimum.

But, since you put the restrictions on the function that both numbers had to be positive, this answer becomes wrong and suddenly all this work is meaningless...

Except now we know that the answer isnt something that we merely solve for algebraically, but will have to induce, logically, as a restriction on a given set of possible answers.

The minimum set of values whose difference is 6 is:
0 and 6.

Their product is zero.... cant get any more minimal than that without going negative...And going negative is a mathematical impossibility if both numbers are positive.

But, both numbers are not positive... they are, however, non-negative (zero isnt positive or negative)

So, if you want two positive numbers then you would have to find the set of numbers that are infinitesimally larger than 0 and 6, but whose difference is still six.

In other words, the answer is the set: 0 + z and 6 + z, where z is an infinitesimally small value.

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tsr21 is correct about his analysis. The answers will have the limits 0 and 6.... which is what I indicated by saying that z, above, has an undefined infinitesimal value.

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Helmut is also correct, right up until he said "x > 0". In fact, I liked his approach and commend him on his method.

However, he made a logical fallacy when he jumped from x > 0 to x = 1... he completely forgot about all the non-integer values that exist in between 0 and 1. He was right about x > 0... which is also what I intended by 0 + z, where z is infinitesimal. x would be greater than 0 if it were equal to 0 + z.

Answer 3

The desired numbers are 1 and 7, by inspection.
edit:
Note that the requirement is that both numbers must be > 0.
Since you want a quadratic solution,
Let x = the smaller number.
then the larger number is x + 6
and the desired quadratic is
N = x(x + 6) = x^2 + 6x
Putting this in modified vertex form,
N = (x + 3)^2 - 9
N must be positive
(x + 3)^2 > 9
x + 3 > 3
x > 0
x = 1
x + 6 = 7

Answer 4

Let the two positive real numbers be x and y where x is the larger number and y is the smaller number.

Since both numbers are positive, their minimum product will be 0.

x - y = 6
x = y + 6 --- (1)

xy = 0 --- (2)

Solving (1) and (2) simultanously,
y(y + 6) = 0
y = 0 or y = -6

y = -6 rejected becuase y must be positive.

x = 0 + 6
x = 6

Therefore, the two positive real numbers are 0 and 6.

Answer 5

If the two numbers must be positive, then this problem doesn't make much sense because obviously the answer is x1=0+ and x2= 6.
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If the two numbers are just real numbers, then
f(x) = x(x-6)
f'(x) = x-6+x
Solve f'(x) = 0 for x,
x = 3
x-6 = -3
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If you want to use algebra, then from the symmetry, the minimum must occur at x = (0+6)/2 = 3. You have the same answer.
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Answer 6

permit x be the bigger extensive style and y be the smaller extensive style. y = x - 6 permit f(x) be the product. f(x) = xy = x(x - 6) = x^2 - 6x f(x)'s minimal happens the place f '(x) = 0 and f ''(x) > 0. f '(x) = 2x - 6 0 = 2x - 6 x = 3 f ''(x) = 2 > 0 for all x so all of us comprehend minimal happens at x = 3. y = 3 - 6 = -3 the two numbers are 3 and -3.

Answer 7

a - b = 6
a = b + 6

ab = minumum
(b+6)(b) = m

f(x) = b^2 + 6b
find the vertex...
= (b^2 + 6b +9) - 9
=(b+3)^2 -9
vertex (-3, -9)

b = -3
a = 3
product = -9