If you have 3.14 pi and you want to divide it by two what would you get

1) Find two real numbers whose sum is 30 and whose product is maximized.

sladkih [1.3K]

Answer:

1) Both 15.

2) 25 and -25.

Step-by-step explanation:

1) Let the 2 numbers be x and 30 - x.

The product = x(30 - x)

f(x) = x(30 - x)

f(x) = 30x - x^2

Finding the derivative:

f'(x) = 30 - 2x

Finding the maximum:

30 - 2x = 0

x = 15.

This gives a maximum f(x) because f"(x) = -2.

So the numbers are 15 and 30 - 15 = 15.

2). If one number is x the other is y.

x - y = 50

y = x - 50

The product =

x(x - 50)

= x^2 - 50x

Finding the derivative:

2x - 50 = 0 for a minimum value.

2x = 50

x = 25.

So the numbers are 25 and 25-50 = -25.

6 0

4 years ago

Please help me! This is is rational function and I don’t know how to/ don’t remember how do this! How would I find and write the

ivanzaharov [21]

An answer is

$\displaystyle f\left(x\right)=\frac{\left(x+1\right)^3}{\left(x+2\right)^2\left(x-1\right)}$

Explanation:

Template:

$\displaystyle f(x) = a \cdot \frac{(\cdots) \cdots (\cdots)}{( \cdots )\cdots( \cdots )}$

There is a nonzero horizontal asymptote which is the line y = 1. This means two things: (1) the numerator and degree of the rational function have the same degree, and (2) the ratio of the leading coefficients for the numerator and denominator is 1.

The only x-intercept is at x = -1, and around that x-intercept it looks like a cubic graph, a transformed graph of $y = x^3$ ; that is, the zero looks like it has a multiplicty of 3. So we should probably put $(x+1)^3$ in the numerator.

We want the constant to be a = 1 because the ratio of the leading coefficients for the numerator and denominator is 1. If a was different than 1, then the horizontal asymptote would not be y = 1.

So right now, the function should look something like

$\displaystyle f(x) = \frac{(x+1)^3}{( \cdots )\cdots( \cdots )}.$

Observe that there are vertical asymptotes at x = -2 and x = 1. So we need the factors $(x+2)(x-1)$ in the denominator. But clearly those two alone is just a degree-2 polynomial.

We want the numerator and denominator to have the same degree. Our numerator already has degree 3; we would therefore want to put an exponent of 2 on one of those factors so that the degree of the denominator is also 3.

A look at how the function behaves near the vertical asympotes gives us a clue.

Observe for x = -2,

as x approaches x = -2 from the left, the function rises up in the positive y-direction, and
as x approaches x = -2 from the right, the function rises up.

Observe for x = 1,

as x approaches x = 1 from the left, the function goes down into the negative y-direction, and
as x approaches x = 1 from the right, the function rises up into the positive y-direction.

We should probably put the exponent of 2 on the $(x+2)$ factor. This should help preserve the function's sign to the left and right of x = -2 since squaring any real number always results in a positive result.

So now the function looks something like

$\displaystyle f(x) = \frac{(x+1)^3}{(x+2 )^2(x-1)}.$

If you look at the graph, we see that $f(-3) = 2$ . Sure enough

$\displaystyle f(-3) = \frac{(-3+1)^3}{(-3+2 )^2(-3-1)} = \frac{-8}{(1)(-4)} = 2.$

And checking the y-intercept, f(0),

$\displaystyle f(0) = \frac{(0+1)^3}{(0+2 )^2(0-1)} = \frac{1}{4(-1)} = -1/4 = -0.25.$

and checking one more point, f(2),

$\displaystyle f(2) = \frac{(2+1)^3}{(2+2 )^2(2-1)} = \frac{27}{(16)(1)} \approx 1.7$

So this function does seem to match up with the graph. You could try more test points to verify.

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If you're extra paranoid, you can test the general sign of the graph. That is, evaluate f at one point inside each of the key intervals; it should match up with where the graph is. The intervals are divided up by the factors:

x < -2. Pick a point in here and see if the value is positive, because the graph shows f is positive for all x in this interval. We've already tested this: f(-3) = 2 is positive.
-2 < x < -1. Pick a point in here and see if the value is positive, because the graph shows f is positive for all x in this interval.
-1 < x < 1. Pick a point here and see if the value is negative, because the graph shows f is negative for all x in this interval. Already tested since f(0) = -0.25 is negative.
x > 1. See if f is positive in this interval. Already tested since f(2) = 27/16 is positive.

So we need to see if -2 < x < -1 matches up with the graph. We can pick -1.5 as the test point, then

$\displaystyle f(-1.5) = \frac{\left(-1.5+1\right)^3}{\left(-1.5+2\right)^2\left(-1.5-1\right)} = \frac{(-0.5)^3}{(0.5)^2(-2.5)} \\= (-0.5)^3 \cdot \frac{1}{(0.5)^2} \cdot \frac{1}{-2.5}$

We don't care about the exact value, just the sign of the result.

Since $(-0.5)^3$ is negative, $(0.5)^2$ is positive, and $(-2.5)$ is negative, we really have a negative times a positive times a negative. Doing the first two multiplications first, (-) * (+) = (-) so we are left with a negative times a negative, which is positive. Therefore, f(-1.5) is positive.