Find the equation of the regression line for the given points. Round any final values to the nearest hundredth, if necessary.
\[(5.6, 8.8), (6.3, 12.4), (7, 14.8), (7.7, 18.2), (8.4, 20.8)\]
The regression line can be modeled as \(y=mx+b\). Given the points above, we can solve for \(m\) and \(b\) by using the following equations:
\[m = \frac{n \sum_{i=1}^{n} x_i y_i - \sum_{i=1}^{n} x_i \sum_{i=1}^{n}y_i}{n \sum_{i=1}^{n}x_i^2 - (\sum_{i=1}^{n}x_i)^2}\]
\[b = \frac{\sum_{i=1}^{n}y_i - m \sum_{i=1}^{n} x_i}{n}\]
df <- data.frame("x" = c(5.6,6.3,7,7.7,8.4), "y" = c(8.8,12.4,14.8,18.2,20.8))
df$xy <- df$x*df$y
df$x2 <- df$x * df$x
n <- nrow(df)
sumx <- sum(df$x)
sumy <- sum(df$y)
sumxy <- sum(df$xy)
sumx2 <- sum(df$x2)
m <- ((n * sumxy) - (sumx * sumy))/((n * sumx2) - (sumx)^2)
b <- (sumy - (m*sumx) )/ n
round(m,2)## [1] 4.26round(b,2)## [1] -14.8Therefore, we know our final equation is: \[y = 4.26x-14.8\]
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. \[f(x,y) = 24x - 6xy^2 - 8y^3\]
First, identify the derivatives with respect to \(x\) and \(y\): \[F_x(x,y) = 24 - 6y^2\] \[F_y(x,y) = -12xy-24y^2\]
Find the critical points of \(y\) by setting \(F_x(x,y) = 0\) and solving for \(y\). \[F_x(x,y) = 24 - 6y^2\] \[6y^2 = 24 \] \[y^2 = 4 \] \[y=2, y=-2\] Find the critical points of \(x\) by subbing \(y=2\) and \(y=-2\) into \(F_y(x,y)\): \[F_y(x,y) = -12xy-24y^2\] For \(y=2\): \[F_y(x,2)= -12x(2)-24(2^2)\] \[24x=-96\] \[x=-4\] For \(y=-2\): \[F_y(x,-2)= -12x(-2)-24(-2^2)\] \[-24x=-96\] \[x=4\]
Another critical point is \((0,0)\) because this also solves \(F_y(x,y)\): \[F_y(x,y) = -12xy-24y^2\] \[ 12xy= -24y^2\] \[ xy= -2y^2\] \[ x= -2y\]
Next, using \(F_x(x,y)\) and \(F_y(x,y)\), find the second derivatives in terms of \(F_{xx}(x,y)\), \(F_{yy}(x,y)\), and \(F_{xy}(x,y)\):
\[F_{xx}(x,y) = 0\] \[F_{yy}(x,y) = -12x - 48y\] \[F_{xy}(x,y) = -12y\] Use the second derivative test provided the rules defined:
\(D = D(x,y) = F_{xx}(x,y) \cdot F_{yy}(x,y) - (F_{xy}(x,y))^2\)
\(D >0\) and \(F_{xx}(x,y) > 0\), then \(F(x,y)\) is a local minimum.
\(D >0\) and \(F_{xx}(x,y) < 0\), then \(F(x,y)\) is a local maximum.
\(D < 0\) then \(F(x,y)\) is a saddle point.
The critical points are \((-4,2)\) , \((4,-2)\), and \((0,0)\) and will be used to solve for \(D\):
At \((-4,2)\):
x <- -4
y <- 2
fxx <- 0
fyy <- (-12*x) - (48*y)
fxy <- -12*y
d <- (fxx * fyy) - (fxy)^2
d## [1] -576At \((4,-2)\):
x <- 4
y <- -2
fxx <- 0
fyy <- (-12*x) - (48*y)
fxy <- -12*y
d <- (fxx * fyy) - (fxy)^2
d## [1] -576At \((0,0)\):
x <- 0
y <- 0
fxx <- 0
fyy <- (-12*x) - (48*y)
fxy <- -12*y
d <- (fxx * fyy) - (fxy)^2
d## [1] 0This tells us that we have 2 saddle points: \((-4,2)\) and \((4,-2)\). We cannot make a conclusion about \((0,0)\) using the second derivatives test.
x <- -4
y <- 2
z <- 24*x−6*x*(y^2)−8*(y^3)
z## [1] -64x <- 4
y <- -2
z <- 24*x−6*x*(y^2)−8*(y^3)
z ## [1] 64Final solution:
\[(-4,2,-64),(4,-2,64)\]
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?
Revenue = Number of house brand units * cost of house brand + Number of name brand units * cost of name brand
\[R(x,y) = (81 - 21x + 17y)x + (40 + 11x - 23y)y\] \[R(x,y) = 81x - 21x^2 + 17yx + 40y + 11xy - 23y^2\] \[R(x,y) = 81x - 21x^2 + 28yx + 40y - 23y^2\]
x <- 2.30
y <- 4.10
r <- 81*x - 21*x^2 + 28*y*x + 40*y - 23*y^2
r## [1] 116.62A 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?
From the problem statement, we know that \(x+y = 96\), or \(x = 96-y\). We will use this to rewrite our function: \[C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700\] \[C(y) = \frac{1}{6}(96-y)^2 + \frac{1}{6}y^2 + 7(96-y) + 25y + 700\] \[C(y) = \frac{1}{6}(y^2-192y+9216) + \frac{1}{6}y^2 - 7y+ 672 + 25y + 700\]
\[C(y) = \frac{1}{6}y^2- 32y +1536 + \frac{1}{6}y^2 +18y + 1372\]
\[C(y) = \frac{1}{3}y^2 - 14y + 2908\]
To find the minimum value, we will take the derivative of \(C(y)\) and set equal to \(0\) to solve for \(y\).
\[C'(y) = \frac{2}{3}y - 14\] \[0 = \frac{2}{3}y - 14\] \[\frac{2}{3}y = 14\] \[y = 21\]
We can plug this back into our original equation \(x = 96-y\) to get the value of \(x\):
\[x = 96-21 = 75\]
Our final values are: \((75, 21)\).
Evaluate the double integral on the given region. \[\int \int_{R}(e^{8x+3y})dA\] \(R:2 \leq x \leq 4\) and \(2 \leq y \leq 4\)
Write your answer in exact form without decimals.
\[\int_2^4 \int_2^4(e^{8x+3y})dxdy\] \[\frac{1}{8} \int_2^4 e^{8x+3y}\Big|_2^4 dy\] \[\frac{1}{8} \int_2^4 (e^{32+3y} - e^{16+3y}) dy\] \[\frac{1}{8} \cdot \frac{1}{3} \Big(\int_2^4 e^{32+3y}dy - \int_2^4e^{16+3y} dy\Big)\] \[\frac{1}{24} \Big(e^{32+3y}\Big|_2^4 - e^{16+3y}\Big |_2^4 \Big)\] \[\frac{1}{24} \Big(e^{44} - e^{38} - e^{28} - e^{22} \Big)\]