Let [tex]\( C(x) \)[/tex] be the cost to produce [tex]\( x \)[/tex] batches of widgets, and let [tex]\( R(x) \)[/tex] be the revenue in thousands of dollars. Complete parts (a) through (d) below.

[tex]\[ R(x) = -x^2 + 6x \][/tex]
[tex]\[ C(x) = x + 4 \][/tex]

(a) Identify the vertex of [tex]\( R(x) \)[/tex].

The vertex of [tex]\( R(x) \)[/tex] is at [tex]\((3, 9)\)[/tex].
(Type an ordered pair. Simplify your answer.)

(b) Identify the y-intercept of [tex]\( C(x) \)[/tex].

The y-intercept of [tex]\( C(x) \)[/tex] is at [tex]\( y = 4 \)[/tex].
(Simplify your answer.)

(c) Graph both functions. Use the graphing tool to graph the functions.
[tex]\[ \square \][/tex]

(d) Find the minimum break-even quantity.

Using the expressions [tex]\(-x^2 + 6x\)[/tex] and/or [tex]\(x + 4\)[/tex], identify an equation to be solved in order to find the minimum break-even quantity.
[tex]\[ \square \][/tex]
(Type an equation.)



Answer :

Certainly! Let's proceed step-by-step to solve through the given parts:

### Part (a)

#### Vertex of [tex]\( R(x) \)[/tex]

First, consider the revenue function:
[tex]\[ R(x) = -x^2 + 6x \][/tex]

This is a quadratic function in standard form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = -1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = 0 \)[/tex].

The vertex of a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex] is at:
[tex]\[ x = -\frac{b}{2a} \][/tex]

Substitute [tex]\( a = -1 \)[/tex] and [tex]\( b = 6 \)[/tex]:
[tex]\[ x = -\frac{6}{2(-1)} = 3 \][/tex]

Now, find the corresponding [tex]\( y \)[/tex]-value (which is [tex]\( R(3) \)[/tex]):
[tex]\[ R(3) = -3^2 + 6(3) = -9 + 18 = 9 \][/tex]

So, the vertex of [tex]\( R(x) \)[/tex] is at:
[tex]\[ (3, 9) \][/tex]

#### y-intercept of [tex]\( C(x) \)[/tex]

Consider the cost function:
[tex]\[ C(x) = x + 4 \][/tex]

The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ C(0) = 0 + 4 = 4 \][/tex]

So, the y-intercept of [tex]\( C(x) \)[/tex] is at [tex]\( y = 4 \)[/tex].

### Part (b)

#### Finding the Minimum Break-even Quantity

To find the break-even quantity, we need to determine the [tex]\( x \)[/tex]-values where the revenue [tex]\( R(x) \)[/tex] equals the cost [tex]\( C(x) \)[/tex]. Therefore, we set
[tex]\[ R(x) = C(x) \][/tex]

Using the given functions:
[tex]\[ -x^2 + 6x = x + 4 \][/tex]

This needs to be solved for [tex]\( x \)[/tex]. Simplify by moving all terms to one side of the equation:
[tex]\[ -x^2 + 6x - x - 4 = 0 \][/tex]
[tex]\[ -x^2 + 5x - 4 = 0 \][/tex]

Now, solve this quadratic equation:
[tex]\[ x^2 - 5x + 4 = 0 \][/tex]

Solve using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 4 \)[/tex]:
[tex]\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{25 - 16}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{9}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm 3}{2} \][/tex]

This gives us two solutions:
[tex]\[ x = \frac{5 + 3}{2} = 4 \][/tex]
[tex]\[ x = \frac{5 - 3}{2} = 1 \][/tex]

Thus, the break-even quantities where the revenue equals the cost are [tex]\( x = 4 \)[/tex] and [tex]\( x = 1 \)[/tex].

However, the minimum break-even quantity is the smallest [tex]\( x \)[/tex] value where this occurs, which is [tex]\( x = 1 \)[/tex].

Therefore, the minimum break-even quantity is:
[tex]\[ \boxed{1} \][/tex]

If any additional parts are required to graph the functions, use the identified critical points and general shapes of the parabola (concave down for [tex]\( R(x) \)[/tex]) and the linear function [tex]\( C(x) \)[/tex]. But primarily, the break-even analysis accurately identifies the minimum value where costs and revenues intersect.