DATA605 Homework 15

1. 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)

xy <- lm(x ~ y)
plot(x ~ y)
abline(xy)

summary(xy)
## 
## Call:
## lm(formula = x ~ y)
## 
## Residuals:
##        1        2        3        4        5 
##  0.05121 -0.09143  0.04681 -0.04901  0.04241 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 3.489004   0.125618   27.77 0.000102 ***
## y           0.234066   0.008061   29.04 8.97e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.0761 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-05

The equation for the regression line for the given points is:

\[\hat{y} = 3.489 = 0.234x\]

2. 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.

\[f(x,y) = 24x - 6xy^2 - 8y^3\] \[\frac{\partial f}{\partial x} = 24 - 6y^2\] \[\frac{\partial f}{\partial y} = -12xy - 24y^2\] \[6y^2 = 24\] \[y^2 = 4\] \[y = +- 2\]

\[-12xy = 24y^2\] \[x = -2y\] \[x = +- 4\] Critical points: (4,-2), (-4,2)

The saddle points of this function are at (x,y) = (-4,2),(4,-2).

3. 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 \(R (x, y)\). \[R(x,y) = x(81-21x + 17y) + y(40 + 11x - 23y)\] \[= 81x - 21x^2 + 17xy + 40y + 11xy - 23y^2\] \[= 81x + 40y + 28xy - 21x^2 - 23y^2\]

Step 2. What is the revenue if she sells the “house” brand for \(2.30\) and the “name” brand for \(4.10\)? \[R(2.3, 4.1) = 81 * 2.3 + 40 * 4.1 + 28 * 2.3 * 4.1 - 21 * (2.3)^2 - 23 * (4.1)^2\] \[= 116.62\] If she sells the “house” brand for \(2.30\) and the “name” brand for \(4.10\), she will make \(116.62\) in revenue.

4. 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?

Consider \(x + y = 96\), then \(x = 96 - y\) \[C(x,y) = C(96-y,y)\] \[= \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700\] \[= \frac{1}{6}(96-y)^2 + \frac{1}{6}y^2 + 7(96-y) + 25y + 700\] \[= \frac{1}{6}(y^2 - 192y + 9216) + \frac{1}{6}y^2 + 672 - 7y + 25y + 700\] \[= \frac{1}{6}y^2 - 32y + 1536 + \frac{1}{6}y^2 _ 18y + 1372\] \[\frac{1}{3}y^2 - 14y + 2908\] \[C_1(y)\] \[C_1(y) = \frac{2}{3}y-14\] \[\frac{2}{3}y - 14 = 0\] \[y = 21\] \[x = 96 - y\] \[x = 75\]

In order to minimize the total weekly cost, the Los Angeles plant should produce 75 units and the Denver plant should produce 21 units.

5. 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.

\[\int_2^4 \int_2^4 e^{8x + 3y} dx dy\] \[\int_2^4 \bigg[\frac{e^{8x+3y}}{8} \bigg]_2^4 \] \[\frac{1}{8} \int_2^4 \big(e^{32 + 3y} - e^{16 + 3y}\big) dy\] \[\frac{1}{8} * \frac{1}{3} \bigg[ \big[e^{32 + 3y}\big]_2^4 - \big[e^{16 + 3y}\big]_2^4 \bigg] dy\] \[\frac{1}{24} \bigg[e^{44} - e^{38}\bigg] - \bigg[e^{28} - e^{22}\bigg]\]