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 )
Using R.
x <- c(5.6,6.3,7,7.7,8.4)
y <- c(8.8,12.4,14.8,18.2,20.8)
reg_df <- data.frame(x,y)
reg_df## x y
## 1 5.6 8.8
## 2 6.3 12.4
## 3 7.0 14.8
## 4 7.7 18.2
## 5 8.4 20.8reg_lm <- lm(y ~ x, reg_df)
reg_lm##
## Call:
## lm(formula = y ~ x, data = reg_df)
##
## Coefficients:
## (Intercept) x
## -14.800 4.257Gives us: \(y = 4.257x - 14.8\)
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.
\(f(x,y) = 24x - 6xy^2 - 8y^3\)
Partial derivatives:
\(f_x(x,y) = 24 - 6y^2\)
\(y = +-2\)
\(f_y(x,y) = -24y^2 - 12xy\)
\(x = -2y\)
Gives us critical points: (-4,2);(4,-2)
Second partial derivatives:
\(f_{xx}(x,y) = 0\)
\(f_{yy}(x,y) = -48y - 12x\)
\(f_{xy}(x,y) = -12y\)
Now D:
\(D = f_{xx} * f_{yy} - f_{xy}^2 = -144y^2\)
Both critical points will be negative here, so both are saddle points.
In (x,y,z) form at the second derivative, they’re (4,2,-64) and (-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 \(R(x,y)\).
\(R(x,y) = x * (81 - 21x + 17y) + y * (40 + 11x - 23y)\)
\(= 81x - 21x^2 + 17xy + 40y - 11xy - 23y^2\)
Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?
81*(2.3) - 21*(2.3)^2 + 17*(2.3)*(4.1) + 40*(4.1) + 11*(2.3)*(4.1) - 23*(4.1)^2## [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?
Let’s find the minimum.
\(C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700\)
If \(x + y = 96\) then \(x = 96 - y\). Substituting:
\(C(y) = \frac{1}{6}(96-y)^2 + \frac{1}{6}y^2 + 7(96 - y) + 25y + 700\)
\(C(y) = \frac{1}{6}(96-y)^2 + \frac{1}{6}y^2 + 7(96 - y) + 25y + 700\)
\(C(y) = \frac{1}{6}y^2 - 32y + 1536 + \frac{1}{6}y^2 + 672 - 7y + 25y + 700\)
\(C(y) = \frac{1}{3}y^2 - 14y + 2908\)
\(C'(y) = \frac{2}{3}y - 14 = 0\)
Putting y at \(14 * \frac{3}{2} = 21\) and therefore \(x = 75\)
Second derivative is positive, making (75,21) the minimum cost point. To minimize cost, LA gets 75; Denver 21 units.
Evaluate the double integral on the given region.
\(\int_2^4 \int_2^4 e^{8x + 3y}dA\);
\(R:2 \leq x \leq 4\)
and \(2 \leq y \leq 4\)
Write your answer in exact form without decimals.
\(\int_2^4 \int_2^4 e^{8x + 3y} dx dy\)
\(\int_2^4 e^{8x} dx \int_2^4 e^{3y} dy\)
\(\frac{1}{8}e^{8x} \vert_2^4 * \frac{1}{3}e^{3y} \vert_2^4\)
\(\frac{1}{24}(e^{32} - e^{16})(e^{12} - e^6)\)
Sources Reviewed:
https://tutorial.math.lamar.edu/classes/calciii/relativeextrema.aspx