A line has an y-intercept of (0,2) and a slope of 3. The point of (a, 11) is also on the line. What

Given:
Cost to build a bookshelf = $20
Cost to build a table = $45
Amount available to spend = $600

Let x = number of bookshelves built.
Let y = number of tables built.

The total number of bookshelves and tables = 18.
Therefore
x + y = 18.
That is,
y = 18 - x (1)

The total amount available to build x bookshelves and y tables = $600. Therefore
20x + 45y = 600
That is (dividing through by 5),
4x + 9y = 120 (2)

Substitute (1) into (2).
4x + 9(18 - x) = 120
4x + 162 - 9x = 120
-5x = -42
x = 8.4
From (1),obtain
y = 18 - 8.4 = 9.6

Because we cannot have fractional bookshelves and tables, we shall test values of x=8, 9 and y=9,10 for profit

Note: The profit is $60 per bookshelf and $100 per table.
If x = 8, then y = 18-8 = 10.
The profit = 8*60 + 10*100 = $1480

If x = 9, then y = 18-9 = 9.
The profit = 9*60 + 9*100 = $1440

The choice of 8 bookshelves and 10 tables is more profitable.

Answer: 8 bookshelves and 10 tables.

5 0

3 years ago

Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh

Vedmedyk [2.9K]

Answer:

Increasing: $x$ and $x>1$ .

Decreasing: $-1$

Step-by-step explanation:

We have been given a function $f(x)=x^3-3x+4$ . We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

$f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)$

$f'(x)=3x^{3-1}-3+0$

$f'(x)=3x^{2}-3$

Let us equate derivative with 0 as find critical points as:

$0=3x^{2}-3$

$3x^{2}=3$

Divide both sides by 3:

$x^{2}=1$

Now we will take square-root of both sides as:

$\sqrt{x^{2}}=\pm\sqrt{1}$

$x=\pm 1$

$x=-1,1$

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

$f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9$