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 )
We can use R to find the regression line by putting the points into a matrix and fitting a regression model on the data.
x <- c(5.6,6.3,7,7.7,8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
lm_model <- lm(y~x)
intercept <- round(coef(lm_model)[1], 2)
slope <- round(coef(lm_model)[2], 2)
# Print the equation of the regression line
cat("y =", slope, "*x +", intercept)## y = 4.26 *x + -14.8Find 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 find the partial derivatives:
\(f_x = 24-6y^2\)
\(f_y=-12xy - 24y^2\)
Now set the partial derivatives equal to 0 to get the critial points:
\(24-6y^2=0\)
\(24=6y^2\)
\(y= \pm 2\)
Now plug into the other equation to get the x values:
For y = 2
\(-12x(2) - 24(2)^2=0\)
\(-24x-96=0\)
\(x = -4\)
For y = -2
\(-12x(-2) - 24(-2)^2=0\)
\(24x - 96 = 0\)
\(x = 4\)
So the two critical points are (4,-2) and (-4, 2)
Now to find out what kind of critial points they are find the second partial derivatives:
\(f_{xx}=0\)
\(f_{xy}=-12y\)
\(f_{yy} = -12x - 48y\)
\(f_{yx} = -12y\)
Now construct the Hessian matrix and compute its determinant:
\[ \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \\ \end{bmatrix} \]
So we get:
\[ \begin{bmatrix} 0 & -12y \\ -12y & -12x-48y \\ \end{bmatrix} \]
The determinant is equal to: \((0)(-12x-48y)-(-12y)(-12y)= -144y^2\)
Now the point (-4,2):
\(-144(2)^2=-576\)
Since its negative this is a saddle point.
Now the point (-4,2):
\(-144(2)^2=-576\)
Since this one is also negative it is also a saddle point.
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 is equal to the number of units sold times the price of each unit
Let x = price of house brand product
Let y = price of name brand product
So, the revenue function is equal to:
\(R(x,y) = x(81-21x+17y) + y(40 + 11x -23y)\)
To find the revenue if the house brand is 2.30 and the name brand is 4.10 we plug into our revenue function
\(R(2.30, 4.10) = 2.30(81-21(2.30)+17(4.10)) + 4.10(40+11(2.30)-23(4.10)) = 116.62\)
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?
x = The number of units sold in LA
y = The number of units sold in Denver
We know that x + y = 96
So then, y = 96 - x
Now we can substitute in for y:
\(C(x) = \frac{1}{6}x^2 + \frac{1}{6}(96-x)^2+7x+25(96-x) + 700\)
\(C(x)= \frac{1}{6}x^2 + \frac{1}{6}(9216 - 162x + x^2) -18x + 3100\)
\(C(x) = \frac{1}{3}x^2 - 50x+4636\)
Now to find the minimum find the derivative with respect to x
\(C'(x) = \frac{2}{3}x-50=0\)
\(\frac{2}{3}x=50\)
\(x=75\)
Since x = 75 then y = 96-75 y = 21
To minimize the total weekly cost they should produce 75 units in LA and 21 in Denver
Evaluate the double integral on the given region.
\(\int_2^4 \int_2^4 (e^{8x+3y}) \, dy \, dx\)
First evaluate the inner integral:
\(\int_2^4 e^{8x+3y} dy = 8x \int_2^4 e^{3y} dy\)
\(= \frac{1}{3} e^{3y} |_2^4 = \frac{1}{3} e^{12y} - \frac{1}{3}e^6\)
Now the outer integral:
\(\int_2^4 e^{8x}(\frac{1}{3}e^{12} - \frac{1}{3}e^6) dx\)
\((\frac{1}{3}e^{12} - \frac{1}{3}) \int_2^4 e^{8x} dx = (\frac{1}{3}e^{12} - \frac{1}{3}) (\frac{1}{8}e^{8x})|_2^4 (\frac{1}{3}e^{12} - \frac{1}{3}) (\frac{1}{8}e^{32}-\frac{1}{8}e^{16})\)
\((\frac{1}{3}e^{12} - \frac{1}{3}) (\frac{1}{8}e^{32}-\frac{1}{8}e^{16})= \frac{1}{24}(e^{44}- e^{28}-e^{38}+e^{22})\)