Find the equation of the regression line for the given points. Round any final values to the nearest hundredth, if necessary. ( 5.6, 8.8 ), ( 6.3, 12.4 ), ( 7, 14.8 ), ( 7.7, 18.2 ), ( 8.4, 20.8 )
Using least squares method, the formulas for the slope \(m\) and the y-intercept \(b\) are given by:
\[ m = \frac{n\sum (xy) - \sum x \sum y}{n\sum (x^2) - (\sum x)^2} \] \[ b = \frac{\sum y - m\sum x}{n} \]
Given the points: \((5.6, 8.8), (6.3, 12.4), (7, 14.8), (7.7, 18.2), (8.4, 20.8)\) we get the following sums:
\(\sum x = 5.6 + 6.3 + 7 + 7.7 + 8.4 =
35.0\)
\(\sum y = 8.8 + 12.4 + 14.8 + 18.2 + 20.8 =
75.0\) \(\sum xy = 5.6 \times 8.8 + 6.3
\times 12.4 + 7 \times 14.8 + 7.7 \times 18.2 + 8.4 \times 20.8 =
572.36\)
\(\sum x^2 = 5.6^2 + 6.3^2 + 7^2 + 7.7^2 +
8.4^2 = 250.74\)
Num points \(n = 5\).
Calculate slope and intercept b:
\[ m = \frac{5 \times 572.36 - 35.0 \times 75.0}{5 \times 250.74 - 35.0^2} = 4.26 \] \[ b = \frac{75.0 - 4.26 \times 35.0}{5} = -14.8 \]
\[ y = 4.26x - 14.8 \]
Find all local maxima, local minima, and saddle points for the
function given below. Write your answer(s) in the form ( x, y, z ).
Separate multiple points with a comma.
\[
f( x, y ) = 24x - 6xy^2 - 8y^3
\]
Given the function: \[ f(x, y) = 24x - 6xy^2 - 8y^3 \]
\[ f_x = \frac{\partial}{\partial x}(24x - 6xy^2 - 8y^3) = 24 - 6y^2 \] \[ f_y = \frac{\partial}{\partial y}(24x - 6xy^2 - 8y^3) = -12xy - 24y^2 \]
\(f_x = 0\) and \(f_y = 0\) and solve for \(x\) and \(y\): \[ 24 - 6y^2 = 0 \] \[ y^2 = 4 \] \[ y = \pm 2 \]
\[ -12x(\pm2) - 24(\pm2)^2 = 0 \] \[ -24x - 96 = 0 \] \[ x = -4 \]
The critical points: \[ (-4, 2), (-4, -2) \]
Second partial derivatives: \[ f_{xx} = 0 \] \[ f_{yy} = -12x - 48y \] \[ f_{xy} = -12y \]
\[ H = f_{xx}f_{yy} - (f_{xy})^2 = 0(-12x - 48y) - (-12y)^2 = -144y^2 \]
Using the above at each critical point:Since \(H < 0\) at both critical points, each of these points is a saddle point. There are no local maxima or minima.
The saddle points are: \[ \text{Saddle points: } (-4, 2), (-4, -2) \]
A grocery store sells two brands of a product, the “house” brand and
a “name” brand. The manager estimates that if she sells the “house”
brand for x dollars and the “name” brand for y dollars, she will be able
to sell 81 - 21x + 17y units of the “house” brand and 40 + 11x - 23y
units of the “name” brand.
Step 1. Find the revenue function R ( x, y ).
Step 2. What is the revenue if she sells the “house” brand for $2.30 and
the “name” brand for $4.10?
house brand = \(x\) dollars
name brand = \(y\) dollars
\(R(x, y)\):
\[ \text{# house brand sold} = 81 - 21x + 17y \] \[ \text{# name brand sold} = 40 + 11x - 23y \]
The revenue function \(R(x, y)\) : \[ R(x, y) = x(81 - 21x + 17y) + y(40 + 11x - 23y) \]
\[ = R(x, y) = 81x - 21x^2 + 17xy + 40y + 11xy - 23y^2 \] \[ \textbf{The function}: R(x, y) = -21x^2 + 28xy - 23y^2 + 81x + 40y \]
For prices \(x = \$2.30\) and \(y = \$4.10\), plugging in to the above: \[ R(2.30, 4.10) = -21(2.30)^2 + 28(2.30)(4.10) - 23(4.10)^2 + 81(2.30) + 40(4.10) \]
\[ R(2.30, 4.10) = -21 \times 5.29 + 28 \times 9.43 - 23 \times 16.81 + 186.30 + 164.00 \] \[ R(2.30, 4.10) \approx 116.62 \]
\[ \textbf{The revenue}= $116.62. \]
A company has a plant in Los Angeles and a plant in Denver. The firm is committed to produce a total of 96 units of a product each week. The total weekly cost is given by \(C(x, y) = \frac{1}{6} x^2 + \frac{1}{6} y^2 + 7x + 25y + 700\), where x is the number of units produced in Los Angeles and y is the number of units produced in Denver. How many units should be produced in each plant to minimize the total weekly cost?
Given the cost function: \[ C(x, y) = \frac{1}{6} x^2 + \frac{1}{6} y^2 + 7x + 25y + 700 \]
since \[ y = 96 - x \]
then: \[ x + y = 96 \]
\[ C(x) = \frac{1}{6} x^2 + \frac{1}{6} (96 - x)^2 + 7x + 25(96 - x) + 700 \] \[ C(x) = \frac{1}{3} x^2 - 50x + 4636 \]
\[ C'(x) = \frac{2}{3} x - 50 \] \[ 0 = \frac{2}{3} x - 50 \] \[ x = 75 \]
\[ y = 96 - 75 = 21 \]
\[ C''(x) = \frac{2}{3} \] Since \(C''(x) > 0\), the point \(x = 75\) represents a local minimum.
To minimize the total weekly cost, the company should produce:
75 units in Los Angeles21 units in Denver
Minimum weekly cost: $2761.00.
Evaluate the double integral of the function \(e^{8x + 3y}\) over the region \(R\) defined by \(2 \leq x \leq 4\) and \(2 \leq y \leq 4\).
\[ \int_{2}^{4} \int_{2}^{4} e^{8x + 3y} \, dy \, dx \]
Integrating with respect to \(y\): \[ \int_{2}^{4} e^{8x + 3y} \, dy = e^{8x} \int_{2}^{4} e^{3y} \, dy \] The integral of \(e^{3y}\) is: \[ \int e^{3y} \, dy = \frac{1}{3}e^{3y} + C \] Evaluating from \(y = 2\) to \(y = 4\): \[ \left[ \frac{1}{3}e^{3y} \right]_{2}^{4} = \frac{1}{3}(e^{12} - e^{6}) \]
\[ = e^{8x} \cdot \frac{1}{3}(e^{12} - e^{6}) \]
Integrate wrt \(x\): \[ \int_{2}^{4} e^{8x} \cdot \frac{1}{3}(e^{12} - e^{6}) \, dx = \frac{1}{3}(e^{12} - e^{6}) \int_{2}^{4} e^{8x} \, dx \] The integral of \(e^{8x}\) is: \[ \int e^{8x} \, dx = \frac{1}{8}e^{8x} + C \] Evaluating from \(x = 2\) to \(x = 4\): \[ \left[ \frac{1}{8}e^{8x} \right]_{2}^{4} = \frac{1}{8}(e^{32} - e^{16}) \] \[ = \frac{1}{24}(e^{12} - e^{6})(e^{32} - e^{16}) \] \[ = \frac{1}{24}(e^{44} - e^{28} - e^{38} + e^{22}) \]