\[( 5.6, 8.8 ), ( 6.3, 12.4 ), ( 7, 14.8 ), ( 7.7, 18.2 ), ( 8.4, 20.8 )\]
x <- c(5.6,6.3,7,7.7,8.4)
y <- c(8.8, 12.4,14.8, 18.2,20.8)
df <- data.frame (x,y)
#round the coefficients of the regression line to the nearest hundredth
round(coef(lm(y~x, data=df)),2)
## (Intercept) x
## -14.80 4.26
Regression Line equation: \(y = -14.80 + 4.26x\)
\[f ( x, y ) = 24x - 6xy^2 - 8y^3 \]
Partial derivative : \[f_x = 24 - 6y^2\] \[f_y = -12xy - 24y^2\] Critical points Set \(f_x = 0\)
\[24 - 6y^2 = 0\]
\[6y^2 = 24\] \[y^2 = 4\] \[y = +/- 2\]
Substituting the y values \(f_y = 0\) \[-12xy -24y^2 = 0\] \[xy + 2y^2 = 0\]
Substitute y=2
\[2x + 2*2^2 = 0\]
\[2x + 8 = 0\] \[2x = -8 \] \[x = -4 \]
Substitute y=-2
\[-2x + 8 = 0\] \[2x = 8\] \[x = 4\]
Substitute x & y values (-4,2)
\[f(x,y) = 24x - 6xy^2 - 8y^3\]
\[z= 24(-4) - 6(-4)(2^2) - 8(2^3)\] \[z = -96 + 96 - 64\] \[x = -64\]
Substitute x & y values (4,-2)
\[f(x,y) = 24x - 6xy^2 - 8y^3\] \[z= 24(4) - 6(4)(-2^2) - 8(-2^3)\]
\[z = 96 - 96 + 64\] \[z = 64\]
Revenue = units_sold x price \[R(x,y) = x(81 - 21x + 17y) + y(40 + 11x -23y)\]
\[ = 81x - 21x^2 + 17xy + 40y + 11xy -23y^2\]
\[ = -21x^2 - 23y^2 + 28xy + 81x + 40y\]
\(C(x, y) = \frac{1}{6} x^2 + \frac{1}{6} y^2 + 7x + 25y + 700\), ### 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: x + y = 96 y = 96 - x
\[C(x, y) = \frac{1}{6} x^2 + \frac{1}{6} y^2 + 7x + 25y + 700\] \[ = \frac{1}{6} x^2 + \frac{1}{6} (96 - x)^{2} + 7x + 25(96 - x) + 700\] \[ = \frac{1}{6} x^2 + \frac{1}{6} (9216 - 192x + x^2) + 7x + 2400 - 25x + 700\]
\[ = \frac{1}{6} x^2 + 1536 + 32x + \frac{x^2}{6} + 7x + 2400 - 25x + 700\] \[ = \frac{1}{6} x^2 + 1536 + 32x + \frac{x^2}{6} + 7x + 2400 - 25x + 700\]
\[ = \frac{1}{3} x^2 - 50x + 4636\]
Setting the derivative to zero
\[C' = \frac{2}{3}x - 50\] \[0 = \frac{2}{3}x - 50\] \[\frac{2}{3}x = 50\] \[x = (50 * 3)/2 = 75\]
The number of units produced in LA to minimize the weekly cost is 75 units.
y = 96 - 75 = 21
\[\int \int (e^{8x + 3y})dA; R:2 \le x \le 4, 2 \le y \le 4 \] Write your answer in exact form without decimals.
\(u = 8x + 3y, du = 8\)
\[\frac{1}{8} \int^4_2 e^u du\] \[= \frac {e^{8x + 3y}}{8}^4_2 \]
\[= \frac {e^{8*4 + 3y}}{8} - \frac{e^8(2) + 3y}{8} \] \[= \frac {e^{32 + 3y}}{8} - \frac{e^{16} + 3y}{8} \] \[ = \int^4_2 \frac{e^{32 + 3y} - e^{16 + 3y} }{8} dy\] \[ = \frac{1}{8} \int^4_2 e^{32 + 3y} dy - \frac{1}{8} \int^4_2 e^{16 + 3y} dy\] \[ = (\frac{1}{8}) \frac{e^{32 + 3y}}{3}^4_2 - (\frac{1}{8}) \frac{e^{16 + 3y}}{3}^4_2 \]
\[ = (\frac{1}{8}) (\frac{e^{32 + 3(4)}}{3} - \frac{e^{32 + 3(2)}}{3}) - (\frac{1}{8}) (\frac{e^{16 + 3(4)}}{3} - \frac{e^{16 + 3(2)}}{3})\]
\[ = (\frac{1}{8}) (\frac{e^{32 + 12}}{3} - \frac{e^{32 + 6}}{3}) - (\frac{1}{8}) (\frac{e^{16 + 12}}{3} - \frac{e^{16 + 6}}{3})\] \[ = \frac{e^{44}}{24} - \frac{e^{38}}{24} - \frac{e^{28}}{24} - \frac{e^{22}}{24} \]
\[ = \frac{e^{44} - e^{38} - e^{28} + e^{22}}{24}\]
(1/24)* (exp(1)^44 - exp(1)^38 - exp(1)^28 - exp(1)^22)
## [1] 5.341559e+17