Let's solve the problem step by step: ### **Given:** - Revenue function: \( R(x) = 20x - x^2 \) - Cost function: \( C(x) = 10x + 800 \) - \( \leq x \leq 15 \) We need to **find the break-even points** (where revenue equals cost). --- ## **Step 1: Set Revenue Equal to Cost** \[ R(x) = C(x) \] \[ 20x - x^2 = 10x + 800 \] --- ## **Step 2: Rearrange the Equation** Move all terms to one side: \[ 20x - x^2 - 10x - 800 = \] \[ - x^2 + 10x - 800 = \] Multiply both sides by \(-1\): \[ x^2 - 10x + 800 = \] --- ## **Step 3: Solve the Quadratic Equation** This is of the form \( ax^2 + bx + c = \): \[ x^2 - 10x + 800 = \] Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -10 \), \( c = 800 \): \[ x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 800}}{2 \cdot 1} \] \[ x = \frac{10 \pm \sqrt{100 - 320}}{2} \] \[ x = \frac{10 \pm \sqrt{-310}}{2} \] \[ \sqrt{-310} = i\sqrt{310} \] --- ## **Step 4: Interpret the Result** The discriminant is **negative** (\(-310\)), so the equation has **no real solutions**. This means **the revenue never equals the cost** in the given interval (\( \leq x \leq 15\)), so there is **no break-even point**. --- ## **Final Answer** \[ \boxed{\text{No real break-even point; revenue never equals cost in the given range.}} \] **Explanation:** The quadratic equation yields imaginary solutions, indicating that the break-even point does not occur for any real value of \(x\) within the range \( \leq x \leq 15\).

# Step-by-Step Solution ## **Given:** - Revenue function: \( R(x) = 20x - x^2 \) - Cost function: \( C(x) = 10x + 800 \) - Interval: \( 0 \leq x \leq 15 \) We need to **find the break-even points** where revenue equals cost. --- ## **Step 1: Set Revenue Equal to Cost** Set the revenue equal to the cost: \[ R(x) = C(x) \] Substituting the functions: \[ 20x - x^2 = 10x + 800 \] --- ## **Step 2: Rearrange the Equation** Move all terms to one side: \[ 20x - x^2 - 10x - 800 = 0 \] Simplify the equation: \[ -x^2 + 10x - 800 = 0 \] Multiply both sides by \(-1\): \[ x^2 - 10x + 800 = 0 \] --- ## **Step 3: Solve the Quadratic Equation** This is a quadratic equation in the form \( ax^2 + bx + c = 0 \): - \( a = 1 \) - \( b = -10 \) - \( c = 800 \) Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting the values: \[ x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 800}}{2 \cdot 1} \] Calculate the discriminant: \[ x = \frac{10 \pm \sqrt{100 - 3200}}{2} \] \[ x = \frac{10 \pm \sqrt{-3100}}{2} \] The square root of a negative number: \[ \sqrt{-3100} = i\sqrt{3100} \] --- ## **Step 4: Interpret the Result** The discriminant is **negative** (\(-3100\)), indicating that the equation has **no real solutions**. This means that **the revenue never equals the cost** in the given interval \( 0 \leq x \leq 15\). --- ## **Final Answer** \[ \boxed{\text{No real break-even point; revenue never equals cost in the given range.}} \] ### **Explanation:** The quadratic equation yields imaginary solutions, indicating that the break-even point does not occur for any real value of \(x\) within the specified interval.