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Sagot :
To solve this problem, we need to find the selling price [tex]\( x \)[/tex] that maximizes the profit [tex]\( y \)[/tex] for the company. The equation given to us is:
[tex]\[ y = -10x^2 + 692x - 6342 \][/tex]
This is a quadratic equation of the form [tex]\( y = ax^2 + bx + c \)[/tex], where [tex]\( a = -10 \)[/tex], [tex]\( b = 692 \)[/tex], and [tex]\( c = -6342 \)[/tex].
Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a = -10 \)[/tex]) is negative, the parabola opens downwards, meaning it has a maximum point at the vertex.
To find the value of [tex]\( x \)[/tex] that gives us the maximum profit, we need to find the x-coordinate of the vertex of the parabola. The formula to find the x-coordinate of the vertex for a quadratic equation [tex]\( ax^2 + bx + c \)[/tex] is:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = -10 \)[/tex] and [tex]\( b = 692 \)[/tex]. Plugging in these values, we get:
[tex]\[ x = -\frac{692}{2 \times -10} \][/tex]
[tex]\[ x = -\frac{692}{-20} \][/tex]
[tex]\[ x = \frac{692}{20} \][/tex]
[tex]\[ x = 34.6 \][/tex]
So, the selling price that maximizes the profit is [tex]$34.60. The maximum profit can be found by substituting \( x = 34.6 \) back into the original profit equation. However, since this question only requests the price \( x \) at which the company makes the maximum profit, we do not need to calculate \( y \). Therefore, the widgets should be sold for $[/tex]34.60 to maximize the company's profit.
[tex]\[ y = -10x^2 + 692x - 6342 \][/tex]
This is a quadratic equation of the form [tex]\( y = ax^2 + bx + c \)[/tex], where [tex]\( a = -10 \)[/tex], [tex]\( b = 692 \)[/tex], and [tex]\( c = -6342 \)[/tex].
Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a = -10 \)[/tex]) is negative, the parabola opens downwards, meaning it has a maximum point at the vertex.
To find the value of [tex]\( x \)[/tex] that gives us the maximum profit, we need to find the x-coordinate of the vertex of the parabola. The formula to find the x-coordinate of the vertex for a quadratic equation [tex]\( ax^2 + bx + c \)[/tex] is:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = -10 \)[/tex] and [tex]\( b = 692 \)[/tex]. Plugging in these values, we get:
[tex]\[ x = -\frac{692}{2 \times -10} \][/tex]
[tex]\[ x = -\frac{692}{-20} \][/tex]
[tex]\[ x = \frac{692}{20} \][/tex]
[tex]\[ x = 34.6 \][/tex]
So, the selling price that maximizes the profit is [tex]$34.60. The maximum profit can be found by substituting \( x = 34.6 \) back into the original profit equation. However, since this question only requests the price \( x \) at which the company makes the maximum profit, we do not need to calculate \( y \). Therefore, the widgets should be sold for $[/tex]34.60 to maximize the company's profit.
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