To solve for the number of DVD/Blu-ray players produced given the marginal cost equation [tex]\( C = 0.03x^2 - 4x + 800 \)[/tex] and the marginal cost [tex]\( C = 3057 \)[/tex], we need to set up and solve the equation:
[tex]\[ 0.03x^2 - 4x + 800 = 3057 \][/tex]
First, we rearrange the equation to set it equal to zero:
[tex]\[ 0.03x^2 - 4x + 800 - 3057 = 0 \][/tex]
Simplifying this, we get:
[tex]\[ 0.03x^2 - 4x - 2257 = 0 \][/tex]
This is a quadratic equation in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 0.03 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = -2257 \)[/tex].
To find the roots of this quadratic equation, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(0.03)(-2257)}}{2(0.03)} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{16 + 270.84}}{0.06} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{286.84}}{0.06} \][/tex]
Solving this, we get two solutions for [tex]\( x \)[/tex]:
[tex]\[ x_1 \approx \frac{4 + 16.92}{0.06} \approx 348.939190617861 \][/tex]
[tex]\[ x_2 \approx \frac{4 - 16.92}{0.06} \approx -215.605857284528 \][/tex]
Since the number of DVD/Blu-ray players produced cannot be negative, we discard the negative solution.
Thus, the number of DVD/Blu-ray players produced is approximately [tex]\( x \approx 348.939190617861 \)[/tex].
Hence, the correct answer is:
[tex]\[ 348.939190617861 \text{ DVD/Blu-ray players} \][/tex]
