Answer :
To find the selling price [tex]\( x \)[/tex] that generates a [tex]$50 daily profit, we need to solve for \( x \) in the quadratic function \( y = -10x^2 + 160x - 430 \) when \( y = 50 \).
Step-by-Step Solution:
### 1. Write the Quadratic Equation:
The original quadratic function is given by:
\[ y = -10x^2 + 160x - 430 \]
### 2. Substitute the Given Profit:
We need to find the value of \( x \) that makes the profit \( y = 50 \):
\[ 50 = -10x^2 + 160x - 430 \]
### 3. Rearrange the Equation to Standard Form:
Move the profit value (\$[/tex]50) to the other side of the equation to set it to zero:
[tex]\[ -10x^2 + 160x - 430 - 50 = 0 \][/tex]
[tex]\[ -10x^2 + 160x - 480 = 0 \][/tex]
Now, the quadratic equation is in standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = -10 \)[/tex], [tex]\( b = 160 \)[/tex], and [tex]\( c = -480 \)[/tex].
### 4. Identify the Quadratic Formula:
The standard method to solve a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
### 5. Calculate the Discriminant:
First, we calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute [tex]\( a = -10 \)[/tex], [tex]\( b = 160 \)[/tex], and [tex]\( c = -480 \)[/tex] into the discriminant formula:
[tex]\[ \Delta = 160^2 - 4(-10)(-480) \][/tex]
[tex]\[ \Delta = 25600 - 19200 \][/tex]
[tex]\[ \Delta = 6400 \][/tex]
### 6. Evaluate the Square Root of the Discriminant:
Next, calculate the square root of [tex]\( \Delta \)[/tex]:
[tex]\[ \sqrt{\Delta} = \sqrt{6400} = 80 \][/tex]
### 7. Apply the Quadratic Formula:
Now, use the quadratic formula to find the roots:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x = \frac{-160 \pm 80}{2(-10)} \][/tex]
### 8. Calculate the Two Solutions:
Using the positive and negative values of the square root:
[tex]\[ x_1 = \frac{-160 + 80}{-20} = \frac{-80}{-20} = 4 \][/tex]
[tex]\[ x_2 = \frac{-160 - 80}{-20} = \frac{-240}{-20} = 12 \][/tex]
### Conclusion:
The selling prices that would generate a \$50 daily profit are:
[tex]\[ x = 4 \, \text{dollars} \][/tex]
and
[tex]\[ x = 12 \, \text{dollars} \][/tex].
These steps guide us through solving the quadratic equation for the prices that yield the given profit, ensuring all necessary calculations and justifications are clearly laid out.
[tex]\[ -10x^2 + 160x - 430 - 50 = 0 \][/tex]
[tex]\[ -10x^2 + 160x - 480 = 0 \][/tex]
Now, the quadratic equation is in standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = -10 \)[/tex], [tex]\( b = 160 \)[/tex], and [tex]\( c = -480 \)[/tex].
### 4. Identify the Quadratic Formula:
The standard method to solve a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
### 5. Calculate the Discriminant:
First, we calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute [tex]\( a = -10 \)[/tex], [tex]\( b = 160 \)[/tex], and [tex]\( c = -480 \)[/tex] into the discriminant formula:
[tex]\[ \Delta = 160^2 - 4(-10)(-480) \][/tex]
[tex]\[ \Delta = 25600 - 19200 \][/tex]
[tex]\[ \Delta = 6400 \][/tex]
### 6. Evaluate the Square Root of the Discriminant:
Next, calculate the square root of [tex]\( \Delta \)[/tex]:
[tex]\[ \sqrt{\Delta} = \sqrt{6400} = 80 \][/tex]
### 7. Apply the Quadratic Formula:
Now, use the quadratic formula to find the roots:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x = \frac{-160 \pm 80}{2(-10)} \][/tex]
### 8. Calculate the Two Solutions:
Using the positive and negative values of the square root:
[tex]\[ x_1 = \frac{-160 + 80}{-20} = \frac{-80}{-20} = 4 \][/tex]
[tex]\[ x_2 = \frac{-160 - 80}{-20} = \frac{-240}{-20} = 12 \][/tex]
### Conclusion:
The selling prices that would generate a \$50 daily profit are:
[tex]\[ x = 4 \, \text{dollars} \][/tex]
and
[tex]\[ x = 12 \, \text{dollars} \][/tex].
These steps guide us through solving the quadratic equation for the prices that yield the given profit, ensuring all necessary calculations and justifications are clearly laid out.