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To minimize the labor cost function \(L(x, y) = \frac{3}{2} x^2 + y^2 - 4x - 6y - 2xy + 116\), we need to find the critical points by taking the partial derivatives of \(L\) with respect to \(x\) and \(y\), and setting them equal to zero.
First, we compute the partial derivative of \(L\) with respect to \(x\):
[tex]\[ L_x = \frac{\partial}{\partial x} \left( \frac{3}{2} x^2 + y^2 - 4x - 6y - 2xy + 116 \right) \][/tex]
Calculating each term separately, we get:
[tex]\[ \frac{\partial}{\partial x} \left( \frac{3}{2}x^2 \right) = 3x \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( y^2 \right) = 0 \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( -4x \right) = -4 \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( -6y \right) = 0 \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( -2xy \right) = -2y \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( 116 \right) = 0 \][/tex]
Combining these, we have:
[tex]\[ L_x = 3x - 4 - 2y \][/tex]
Next, we compute the partial derivative of \(L\) with respect to \(y\):
[tex]\[ L_y = \frac{\partial}{\partial y} \left( \frac{3}{2} x^2 + y^2 - 4x - 6y - 2xy + 116 \right) \][/tex]
Calculating each term separately, we get:
[tex]\[ \frac{\partial}{\partial y} \left( \frac{3}{2}x^2 \right) = 0 \][/tex]
[tex]\[ \frac{\partial}{\partial y} \left( y^2 \right) = 2y \][/tex]
[tex]\[ \frac{\partial}{\partial y} \left( -4x \right) = 0 \][/tex]
[tex]\[ \frac{\partial}{\partial y} \left( -6y \right) = -6 \][/tex]
[tex]\[ \frac{\partial}{\partial y} \left( -2xy \right) = -2x \][/tex]
[tex]\[ \frac{\partial}{\partial y} \left( 116 \right) = 0 \][/tex]
Combining these, we have:
[tex]\[ L_y = 2y - 6 - 2x \][/tex]
To find the critical points, we set \(L_x = 0\) and \(L_y = 0\):
[tex]\[ 3x - 4 - 2y = 0 \quad \text{(1)} \][/tex]
[tex]\[ 2y - 6 - 2x = 0 \quad \text{(2)} \][/tex]
First, let's solve equation (2) for \(y\):
[tex]\[ 2y - 6 - 2x = 0 \][/tex]
[tex]\[ 2y = 2x + 6 \][/tex]
[tex]\[ y = x + 3 \][/tex]
Now, substitute \(y = x + 3\) into equation (1):
[tex]\[ 3x - 4 - 2(x + 3) = 0 \][/tex]
[tex]\[ 3x - 4 - 2x - 6 = 0 \][/tex]
[tex]\[ x - 10 = 0 \][/tex]
[tex]\[ x = 10 \][/tex]
Using \(x = 10\) in \(y = x + 3\), we get:
[tex]\[ y = 10 + 3 \][/tex]
[tex]\[ y = 13 \][/tex]
Thus, the values of \(x\) and \(y\) that minimize the labor cost are \(x = 10\) hours and \(y = 13\) hours.
Now, we find the minimum labor cost by substituting \(x = 10\) and \(y = 13\) back into the labor cost function \(L(x, y)\):
[tex]\[ L(10, 13) = \frac{3}{2} (10)^2 + (13)^2 - 4(10) - 6(13) - 2(10)(13) + 116 \][/tex]
Calculate step by step:
[tex]\[ \frac{3}{2} (10)^2 = \frac{3}{2} \cdot 100 = 150 \][/tex]
[tex]\[ (13)^2 = 169 \][/tex]
[tex]\[ -4(10) = -40 \][/tex]
[tex]\[ -6(13) = -78 \][/tex]
[tex]\[ -2(10)(13) = -260 \][/tex]
[tex]\[ 116 \][/tex]
Now, combine these results:
[tex]\[ L(10, 13) = 150 + 169 - 40 - 78 - 260 + 116 \][/tex]
Calculate the total:
[tex]\[ L(10, 13) = 57 \][/tex]
Thus, the minimum labor cost is $57.
To summarize, the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that minimize the labor cost are [tex]\(x = 10\)[/tex] and [tex]\(y = 13\)[/tex], and the minimum labor cost is $57.
First, we compute the partial derivative of \(L\) with respect to \(x\):
[tex]\[ L_x = \frac{\partial}{\partial x} \left( \frac{3}{2} x^2 + y^2 - 4x - 6y - 2xy + 116 \right) \][/tex]
Calculating each term separately, we get:
[tex]\[ \frac{\partial}{\partial x} \left( \frac{3}{2}x^2 \right) = 3x \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( y^2 \right) = 0 \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( -4x \right) = -4 \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( -6y \right) = 0 \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( -2xy \right) = -2y \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( 116 \right) = 0 \][/tex]
Combining these, we have:
[tex]\[ L_x = 3x - 4 - 2y \][/tex]
Next, we compute the partial derivative of \(L\) with respect to \(y\):
[tex]\[ L_y = \frac{\partial}{\partial y} \left( \frac{3}{2} x^2 + y^2 - 4x - 6y - 2xy + 116 \right) \][/tex]
Calculating each term separately, we get:
[tex]\[ \frac{\partial}{\partial y} \left( \frac{3}{2}x^2 \right) = 0 \][/tex]
[tex]\[ \frac{\partial}{\partial y} \left( y^2 \right) = 2y \][/tex]
[tex]\[ \frac{\partial}{\partial y} \left( -4x \right) = 0 \][/tex]
[tex]\[ \frac{\partial}{\partial y} \left( -6y \right) = -6 \][/tex]
[tex]\[ \frac{\partial}{\partial y} \left( -2xy \right) = -2x \][/tex]
[tex]\[ \frac{\partial}{\partial y} \left( 116 \right) = 0 \][/tex]
Combining these, we have:
[tex]\[ L_y = 2y - 6 - 2x \][/tex]
To find the critical points, we set \(L_x = 0\) and \(L_y = 0\):
[tex]\[ 3x - 4 - 2y = 0 \quad \text{(1)} \][/tex]
[tex]\[ 2y - 6 - 2x = 0 \quad \text{(2)} \][/tex]
First, let's solve equation (2) for \(y\):
[tex]\[ 2y - 6 - 2x = 0 \][/tex]
[tex]\[ 2y = 2x + 6 \][/tex]
[tex]\[ y = x + 3 \][/tex]
Now, substitute \(y = x + 3\) into equation (1):
[tex]\[ 3x - 4 - 2(x + 3) = 0 \][/tex]
[tex]\[ 3x - 4 - 2x - 6 = 0 \][/tex]
[tex]\[ x - 10 = 0 \][/tex]
[tex]\[ x = 10 \][/tex]
Using \(x = 10\) in \(y = x + 3\), we get:
[tex]\[ y = 10 + 3 \][/tex]
[tex]\[ y = 13 \][/tex]
Thus, the values of \(x\) and \(y\) that minimize the labor cost are \(x = 10\) hours and \(y = 13\) hours.
Now, we find the minimum labor cost by substituting \(x = 10\) and \(y = 13\) back into the labor cost function \(L(x, y)\):
[tex]\[ L(10, 13) = \frac{3}{2} (10)^2 + (13)^2 - 4(10) - 6(13) - 2(10)(13) + 116 \][/tex]
Calculate step by step:
[tex]\[ \frac{3}{2} (10)^2 = \frac{3}{2} \cdot 100 = 150 \][/tex]
[tex]\[ (13)^2 = 169 \][/tex]
[tex]\[ -4(10) = -40 \][/tex]
[tex]\[ -6(13) = -78 \][/tex]
[tex]\[ -2(10)(13) = -260 \][/tex]
[tex]\[ 116 \][/tex]
Now, combine these results:
[tex]\[ L(10, 13) = 150 + 169 - 40 - 78 - 260 + 116 \][/tex]
Calculate the total:
[tex]\[ L(10, 13) = 57 \][/tex]
Thus, the minimum labor cost is $57.
To summarize, the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that minimize the labor cost are [tex]\(x = 10\)[/tex] and [tex]\(y = 13\)[/tex], and the minimum labor cost is $57.
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