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 )
x1 = 5.6
y1 = 8.8
x2 = 6.3
y2 = 12.4
x3 = 7
y3 = 14.8
x4 = 7.7
y4 = 18.2
x5 = 8.4
y5 = 20.8
x_mean = (x1+x2+x3+x4+x5)/5
x_mean## [1] 7y_mean = (y1+y2+y3+y4+y5)/5
y_mean## [1] 15x_y = ((x1*y1) + (x2*y2) +(x3*y3) +(x4*y4) +(x5*y5))/5
x_y## [1] 109.172x_sq = ((x1**2) + (x2**2) + (x3**2) + (x4**2) + (x5**2))/5
x_sq## [1] 49.98m = ((x_mean * y_mean) - x_y)/((x_mean**2)-x_sq)
m## [1] 4.257143b = y_mean - m*x_mean
b## [1] -14.8y = 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.
\(\ f(x, y) = 24x - 6xy^2 - 8y^3\)
\(\ \frac{df}{dx} = 24 - 6y^2\)
\(\ \frac{df}{dy} = -12xy - 24y^2\)
\(\ \frac{d^2f}{dx} = 0\)
\(\ \frac{d^2f}{dy} = -12xy - 48y\)
\(\ \frac{d^2f}{dxdy} = -12y\)
\(\ \frac{df}{dx}\)
\(\ 24-6y^2 = 0\)
\(\ 4-y^2= 0\)
\(\ y^2 = 4\)
\(\ y = 2, -2\)
\(\ \frac{df}{dy}\)
\(\ -12xy - 24y^2 = 0\)
\(\ x = -2y\)
Substituting the y = 2, -2 into the formula, we get (-4, 2) and (4,-2).
Second derivative
We substitute the points (-4, 2) and (4,-2) into each equation
\(\ \frac{d^2f}{dx} = 0\)
\(\ \frac{d^2f}{dy} = -12xy - 48y\)
\(\ (-4, 2) = -48\) $ (4, -2) = 48
\(\ \frac{d^2f}{dxdy} = -12y\)
\(\ (-4, 2) = -24\) \(\ (4, -2) = 24\)
Now we get the first numbers from each derivative and form our first equation, and do the same for the second equation to get the value of the discriminat.
D = 0*48-(24)**2
D## [1] -576D = 0*(-48)-(-24)**2
D## [1] -576The critcal point is (−4,2)and(4,−2). Since D < 0 the critical point is saddle point of f.
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 Function :
\(\ R=x(81−21x+17y)+y(40+11x−23y)\)
\(\ =81x−21x^2+17xy+40y+11xy−23y^2\)
\(\ =−21x^2−23y^2+28xy+81x+40y\)
Partial derivatives
\(\ \frac{dR}{dx} = −42x+28y+81\)
\(\ \frac{dR}{dy} = −46y+28x+40\)
We find the critical points from the equations
\(\ -42x + 28y + 81 = 0\)
\(\ 28x - 46y + 40 = 0\)
\(\ y= -2.4, x = 0.32\)
# plug in the values into our function.
revenue = -21*(2.30)^2 -23*(4.10)^2 + 28*(2.3)*(4.1) + 81*(2.3) + 40*(4.1)
revenue## [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?
\(\ x+y=96\)
\(\ C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700\)
\(\ C'(X)= \frac{1}{3}x+7\)
\(\ 0=\frac{1}{3}x+7\)
\(\ -7 = \frac{1}{3}x\)
$ -21 = x $
\(\ C'(Y)=\frac{1}{3}y+25\)
\(\ 0=\frac{1}{3}y+25\)
\(\ -25 = \frac{1}{3}y\)
\(\ -75 = y\)
Therefore (x,y) = (75,21)
Evaluate the double integral on the given region.
\[\ \int\int (e^{8x + 3y}) dA ; R: 2 \leq x \leq 4 and 2 \leq y \leq 4\]
Write your answer in exact form without decimals.
\[\ \int\int (e^{8x + 3y}) dA ; R: 2 \leq x \leq 4 and 2 \leq y \leq 4\]
\[\ \int\int(e^{8x + 3y})DA\]
\[\ \int\int(e^{8x + 3y})\frac{3dy}{3}dx\]
\[\ \frac{1}{3}\int_2^4\bigg[e^{8x + 3y}\bigg]_2^4dx\]
\[\ \frac{1}{3}\int_2^4(e^{8x + 12}-e^{ex+6})dx\]
\[\ \frac{1}{3}\int_2^4(e^{8x + 12}-e^{ex+6})*\frac{8}{8}dx\]
\[\ \frac{1}{24}\bigg[e^{8x + 3y}\bigg]_2^4\]
\[\ \frac{1}{24}\bigg[e^{44}−e^{38}−(e^{28}−e^{22})\bigg]_2^4\]
\[\ \frac{1}{24}\bigg[e^{44}−e^{38}−e^{28}+e^{22})\bigg]_2^4\]
\[\ =\frac{1}{24}*1\]