\(( 5.6, 8.8 ), ( 6.3, 12.4 ), ( 7, 14.8 ), ( 7.7, 18.2 ), ( 8.4, 20.8 )\)
library(tidyverse)x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
df <- as.data.frame(x, y)# Rounding values to the nearest hundredth
model1 <- round(coef(lm(y~x, data=df)), 3)
model1## (Intercept) x
## -14.800 4.257Equation of Regression Line: \(y=-14.80x+4.26\)
\(( x, y, z )\). Separate multiple points with a comma.
Function: \(f(x,y)=24x-6xy^2-8y^3\)
Find Partial Derivatives of \(f(x)\) and \(f(y)\):
\(f(x) = 24-6y^2\)
\(f(y) = -12xy-24y^2\)
Set \(f_x\) = 0:
\(24-6y^2=0\)
\(24 = 6y^2\)
\(4 = y^2\)
\(y= \pm2\)
Plug in \(y=2\) for \(f(y) = 0\):
\(-12xy-24y^2 = 0\) \(-12x(2)-24(2)^2 = 0\) \(-24x-96=0\) \(-24x=96\) \(x=-4\)
Same for \(y=-2\):
\(-12xy-24y^2 = 0\) \(-12x(-2)-24(-2)^2 = 0\) \(24x-96=0\) \(24x=96\) \(x=4\)
Calculating \(f(x,y)\):
\[ f(4,−2) = 24(4)−6(4)(−2)^2−8(−2)^3 \\ =96−96+64 \\ f(4,−2)=64 \] \[ f(−4,2)=24(−4)−6(−4)(2)^2−8(2)^3 \\ =−96+96−64\\ f(−4,2)=−64 \]
The critical points are \((4,-2, 64)\) and \((-4,2,-64)\)
Using Second Derivative Test to determine the local maxima, local minima, and saddle points:
\[ f_{xx} = 0 \\ f_{yy} = -12x-48y \\ f_{xy} = -12y \]
\[ D(x,y) = f_{xx}f_{yy}-(f_{xy})^2 \\ D = (0)(-12x-48y)-(-12y)^2 \\ D = 0 - 144y^2 \\ D = -144y^2 \]
Answer: Since \(D<0\), we can conclude that the saddle point is at \(P\), which are the critical points \((4,-2, 64)\) and \((-4,2,-64)\).
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 \(( x, y )\).
\[ R(x,y) = (81- 21x + 17y)x + (40 + 11x - 23y)y \\ R(x, y)=81x - 21x^2 + 17xy + 40y +11xy - 23y^2 \\ R(x,y)= 81x -21x^2 + 28xy +40y - 23y^2 \]
Step 2: What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?
Revenue Function from Step 1:
\[ R(x,y)= 81x -21x^2 + 28xy +40y - 23y^2 \]
Input x and y values into revenue function:
x <- 2.3 # House brand
y <- 4.1 # Name brand
revenue <- (81*x)-(21*x^2) + (28*x*y) + (40*y) - (23*y^2)
cat("The revenue if she sells the house and name brands is $",revenue)## The revenue if she sells the house and name brands is $ 116.62product 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?
Total units produced per week: \(x+y=96\)
To find \(x\): \(x=96-y\)
Plug in \(96-y\) into x values of equation and simplifying:
\[ C(x,y) = \frac{1}{6}(96-y)^2 +\frac{1}{6}y^ 2 + 7(96-y) + 25y + 700 \\ C(x,y) = \frac{1}{6}(y^2 - 192y + 9216) + \frac{1}{6}y^ 2 -7y + 672 + 25y + 700 \\ C(x,y) = \frac{1}{6}y^2 -32y + 1536 + \frac{1}{6}y^ 2 - 7y +672 +25y + 700 \\ C(x,y) = \frac{1}{3}y^2 -14y + 2908 \]
Taking the derivative when set to zero to find y, which represents units produced in Denver:
\[ C′(y)=\frac{2}{3}y - 14 = 0 \\ =\frac{2}{3}y = 14 \\ = 2y = 42 \\ y = 21 \]
Input \(y=21\) into \(x=96-y\) to find x, the number of units produced in Los Angeles:
\[ x=96-21 \\ x=75 \]
Answer: 75 units should be produced in Los Angeles and 21 units should be produced in Denver to minimize total weekly costs.
\[ \iint_R { ({ e }^{ 8x+3y })dA}; R: 2\le x \le 4 \space and \space 2 \le y \le 4 \]
Write your answer in exact form without decimals.
\[ \begin{split} \int_2^4\int_2^4 (e^{8x+3y})\ dy\ dx &= \int_2^4 (\frac{1}{3}e^{8x+3y})|_2^4\ dx\\ &= \int_2^4 ((\frac{1}{3}e^{8x+12})-(\frac{1}{3}e^{8x+6}))\ dx\\ &= \int_2^4 \frac{1}{3}e^{8x+6}(e^6-1)\ dx\\ &= \frac{1}{24}e^{8x+6}(e^6-1) |_2^4\\ &= \frac{1}{24}e^{32+6}(e^6-1)-\frac{1}{24}e^{16+6}(e^6-1)\\ &= \frac{1}{24}(e^{44}-e^{38})(e^{28}-e^{22})\\ &= \frac{1}{24}(e^{44} - e^{38} - e^{28} + e^{22}) \end{split} \]
(1/24)*((exp(44) - exp(38))*(exp(28) - exp(22)))## [1] 7.70612e+29