x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
r <- lm(y~x)
r##
## Call:
## lm(formula = y ~ x)
##
## Coefficients:
## (Intercept) x
## -14.800 4.257The regression model is \(y = -14.8 + 4.257x\).
Partial derivatives:
\(f_x(x, y) = 24 - 6y^2\)
\(f_y(x, y) = -12xy - 24y^2\)
\(f_x\) and \(f_y\) are never underfined.
If \(24-6y^2=0\), then \(y^2 = 4\) and \(y = \pm2\).
If \(y=2\) and \(-12xy - 24y^2=0\), then \(-24x = 24\times 4\) and \(x=-4\).
If \(y=-2\) and \(-12xy - 24y^2=0\), then \(24x = 24\times 4\) and \(x=4\).
Calculate \(f(x, y)\).
\(f(4,-2) = 24\times 4 - 6\times4\times (-2)^2 - 8 \times (-2)^3 = 64\)
\(f(-4,2) = 24\times (-4) - 6\times(-4)\times 2^2 - 8 \times 2^3 = -64\)
Two critical points: \((4,-2,64)\) and \((-4, 2, -64)\).
Use Second Derivative test to determine if points are minimum, maximum or saddle.
Second partial derivatives:
\(f_{xx}=0\)
\(f_{yy}=-12x-48y\)
\(f_{xy}=-12y\)
Then \(D(x,y) = f_{xx} f_{yy}-f^2_{xy} = -(-12y)^2 = -144y^2\).
\(D(x,y)<0\) for all \((x, y)\), so per Second Derivative Test, any critical point is a saddle point.
\[ \begin{split} R(x,y) &= (81 - 21x + 17y)x + (40 + 11x - 23y)y \\ &= 81x-21x^2+17xy+40y+11xy-23y^2\\ &=81x+40y+28xy-21x^2-23y^2 \end{split} \]
\(R(2.3, 4.1)=81\times 2.3+40\times 4.1+28\times 2.3\times 4.1-21\times (2.3)^2-23\times (4.1)^2 = 116.62\)
Consider \(x+y=96\), then \(x=96-y\).
\[ \begin{split} C(x,y) = C(96-y,y) &= \frac{1}{6} x^2 + \frac{1}{6} y^2 + 7x + 25y + 700 \\ &=\frac{1}{6} (96-y)^2 + \frac{1}{6} y^2 + 7\times (96-y) + 25y + 700 \\ &=\frac{1}{6}(y^2 - 192 y + 9216) + \frac{1}{6}y^2+672-7y+25y+700\\ &= \frac{1}{6}y^2 - 32y+1536+\frac{1}{6}y^2+18y+1372\\ &= \frac{1}{3}y^2 - 14y + 2908\\ &=C_1(y) \end{split} \]
\(C_1'(y) = \frac{2}{3}y-14\)
To find the minimal value consider \(C_1'(y)=\frac{2}{3}y-14=0\), then \(y=21\). Then \(x=96-y=75\).
There should be 75 units produced in Los Angeles and 21 units produced in Denver.
\[ \begin{split} \int_2^4\int_2^4 (e^{8x+3y})\ dy\ dx &= \int_2^4 (\frac{1}{3}e^{8x+3y})|_2^4\ dx\\ &= \int_2^4 ((\frac{1}{3}e^{8x+12})-(\frac{1}{3}e^{8x+6}))\ dx\\ &= \int_2^4 \frac{1}{3}e^{8x+6}(e^6-1)\ dx\\ &= \frac{1}{24}e^{8x+6}(e^6-1) |_2^4\\ &= \frac{1}{24}e^{32+6}(e^6-1)-\frac{1}{24}e^{16+6}(e^6-1)\\ &= \frac{1}{24}(e^6-1)(e^{38}-e^{22})\\ &= \frac{1}{24}(e^{44} - e^{38} - e^{28} + e^{22}) \end{split} \]