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)
x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
reg <- lm(y~x)
summary(reg)##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## 1 2 3 4 5
## -0.24 0.38 -0.20 0.22 -0.16
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -14.8000 1.0365 -14.28 0.000744 ***
## x 4.2571 0.1466 29.04 8.97e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.3246 on 3 degrees of freedom
## Multiple R-squared: 0.9965, Adjusted R-squared: 0.9953
## F-statistic: 843.1 on 1 and 3 DF, p-value: 8.971e-05The regression line equation would be \(y = 4.2471x - 14.8\)
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. \[f(x, y) = 24x - 6xy^2 - 8y^3\] \(f_x(x, y) = 24 - 6y^2\) and \(f_y(x, y) = 24y^2 - 12xy\)
To get \(x\) and \(y\) we solve the equations with zero
\(24 - 6y^2 = 0\) then \(y = \pm 2\) substitute in the 2nd equation to get the value of \(x\)
When \(y = +2\) then \(24y^2 - 12xy = 0\) then \(x = -4\)
When \(y = -2\) then \(24y^2 - 12xy = 0\) then \(x = 4\)
\(f(-4,2) = 24\times (-4) - 6\times(-4)\times 2^2 - 8 \times 2^3 = -64\)
\(f(4,-2) = 24\times 4 - 6\times4\times (-2)^2 - 8 \times (-2)^3 = 64\)
Critical points \((-4, 2, -64)\) and \((4, -2, 64)\)
Find maxima and minima
\(f_{xx} = 0\) , \(f_{yy} = -12x-48y\), \(f_{x,y} = -12y\)
then \(D(x,y) = f_{xx} f_{yy}-f^2_{xy} = -(-12y)^2 = -144y^2 < 0\) the any critical point is 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?
\[R(x,y) = (81 - 21x + 17y)x + (40 + 11x - 23y)y\] \[81x-21x^2+17xy+40y+11xy-23y^2\]
\[81x+40y+28xy-21x^2-23y^2\] \[R(2.3, 4.1)=81\times 2.3+40\times 4.1+28\times 2.3\times 4.1-21\times (2.3)^2-23\times (4.1)^2 = 116.62\]
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?
let \(x\): number of products in Los Angles, \(y\): number of products produced in Denver.
Then \(x + y = 96\) and \(x = 96-y\)
\[C(x,y) = C(96-y,y) = \frac{1}{6} x^2 + \frac{1}{6} y^2 + 7x + 25y + 700\] \[C_1(y) = \frac{1}{6} (96-y)^2 + \frac{1}{6} y^2 + 7\times (96-y) + 25y + 700\] \[C_1(y) = \frac{1}{6}(y^2 - 192 y + 9216) + \frac{1}{6}y^2+672-7y+25y+700\] \[C_1(y) = \frac{1}{6}y^2 - 32y+1536+\frac{1}{6}y^2+18y+1372\] \[C_1(y) = \frac{1}{3}y^2 - 14y + 2908\] \[C_1'(y) = \frac{2}{3}y-14\]
minimal value when \(C_1'(y) = 0\) then \(\frac{2}{3}y-14 = 0\) then \(y = 21\) and \(x = 96 - 21 = 75\)
There should be 75 units produced in Los Angeles and 21 units produced in Denver.
Evaluate the double integral on the given region.
\[\int\int_R (e^{8x+3y}) dA, R:2\le x\le4\ and\ 2 \le y \le 4\] \[\int_2^4\int_2^4 (e^{8x+3y})\ dy\ dx = \int_2^4 (\frac{1}{3}e^{8x+3y})|_2^4\ dx\]
\[\int_2^4 ((\frac{1}{3}e^{8x+12})-(\frac{1}{3}e^{8x+6}))\ dx\]
\[\int_2^4 \frac{1}{3}e^{8x+6}(e^6-1)\ dx\]
\[\frac{1}{24}e^{8x+6}(e^6-1) |_2^4\]
\[\frac{1}{24}e^{32+6}(e^6-1)-\frac{1}{24}e^{16+6}(e^6-1)\]
\[\frac{1}{24}(e^6-1)(e^{38}-e^{22})\]
\[\frac{1}{24}(e^{44} - e^{38} - e^{28} + e^{22})\]