- 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 )
Step 1:
Section points to x and y variables
x = c( 5.6, 6.3, 7, 7.7, 8.4)
y = c( 8.8, 12.4, 14.8, 18.2, 20.8)Step 2:
Generate regression model
reg_lm = lm(y~x)Step 3:
Formulate the regression line equation.
lm(y~x)##
## Call:
## lm(formula = y ~ x)
##
## Coefficients:
## (Intercept) x
## -14.800 4.257\(y = -14.80 + 4.257 x\)
- 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.
\(f ( x, y ) = 24x -6xy^2 - 8y^3\)
Step 1:
Solve 1st and 2nd Partial derivatives of \(f_x f_y\):
1st
\(f_x = 24 - 6y^2\)
\(f_y = -12xy - 24y^2\)
2nd
\(f_{xy} = -12y\) \(f_{xx} = 0\)
\(f_{yy} = -12x - 48y\)
Step 2:
Determine critical points by solving for \(f_x(x,y) = 0\) and \(f_y(x,y) = 0\)
\(0 = 24 - 6y^2\)
\(y = sqrt(24/6)\)
\(y = +/-2\)
Solve for x where y = 2
\(0 = -12xy - 24y^2\)
\(0 = -12(2)x - 24(2)^2\)
\(0 = -24x - 96\)
\(x = 96/-24 = -4\)
Solve for x where y = -2
\(0 = -12(-2)x - 24(-2)^2\)
\(0 = 24x - 96\)
\(x = 96/24 = 4\)
When substituting these back in the function
equat = function(x,y){
z = 24*x - 6*x*y^2 - 8*y^3
return(c(x,y,z))
}
print(rbind(equat(-4,2),equat(4,-2)))## [,1] [,2] [,3]
## [1,] -4 2 -64
## [2,] 4 -2 64Critical points are \((-4,2), (4,-2)\)
Step 3:
Second derivative test
\(f_{(x,y)} = 24x - 6xy^2 - 8y^3\)
\(\frac{df}{dx} = 24 - 6y^2\)
\(dfdy = -12xy - 24y^2\)
\(dfdx = 24 - 6y^2 = 0 -> 4 - y^2 = 0\)
\(for(4,−2)f(x,y)=24∗4−6∗4∗(−2)2−8(−2)3=64\) \(for(4,−2)f(x,y)=24∗−4−(6∗−4∗(2)2)−8(2)3=−64\)
Saddle point is (4,-2)
- 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 \(R(x,y)\)
Identify house and name
\(R(x) = (81 - 21x + 17y) \times x\)
\(R(y) = (40 + 11x - 23y) \times y\)
Revenue function \(R(x,y)\) by combining \(R(x)\) and \(R(y)\)
\(R(x,y) = R(x) + R(y)\) \(R(x,y) = x(81−21x+17y)+y(40+11x−23y)\) \(R(x,y) = 28xy−23y^2−21x^2+40y+81x\)
Step 2:
What is the revenue if she sells the “house” brand for 2.30 and the “name” brand for $4.10? \(R(x,y)=2.3∗(81−21∗2.3+17∗4.1)+4.1∗(40+11∗2.3−23∗4.1)\)
Revenue is \(116.62\) dollars.
- 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?
Step 1:
Firm is committed produce 96 from x (Units produced in LA) and y (units produced in Denver), \(x+y = 96\)
\(y = 96 - x\)
\(x = 96 - y\)
Step 2:
Rewrite total weekly cost with respect to x,
\(y = 96 -x\) \(C(x) = \frac{1}{6}x^2 + \frac{1}{6}(96-x)^2 + 7x + 25(96-x) +700\)
\(C(x) = \frac{1}{6}x^2 + 1536 - 32x + \frac{1}{6}x^2 + 7x + 2400 - 25x +700\)
\(C(x) = \frac{1}{3}x^2 - 50x + 4636\)
\(\frac{dc}{dx} = \frac{2}{3}x - 50 = 0\)
\(x = 75\)
\(y = 21\)
75 units from LA and 21 units from Denver should be produced to minimize the total weekly cost.
- Evaluate the double integral on the given region.
\(A= \int\int e^{(8x+3y)} dA\) ; \(R: 2\le x \le 4\) and $2 y 4 $
\(A= \int_{2}^{4}\int_{2}^{4} e^{(8x+3y)} dxdy\)
\(A = \int_{2}^{4}e^{8x}dx \times \int_{2}^{4}e^{3y}dy\)
\(A = \frac {1}{8}e^{8x}|_{2}^{4}\times \frac{1}{3}e^{3y}|_{2}^{4}\)
\(A = \frac{1}{24}e^{8x}|_{2}^{4}\times e^{3y}|_{2}^{4}\)
\(A= \frac{1}{24}(e^{32} - e^{16})(e^{12} - e^{6})\)
(1/24)*(exp(1)^32 - exp(1)^16)*exp(1)^12 - exp(1)^6## [1] 5.354833e+17