Problem 1: Regression Line

Find the equation of the regression line for the given points. Round any final values to the nearest hundredth, if necessary.

\[ (5.6, 8.8), (6.3,12.4), (7,14.8), (7.7,18.2), (8.4,20.8) \]

## (Intercept)           x 
##      -14.80        4.26

Equation of Regression Line:

\[ {y} = -14.80 + 4.26x \]

Problem 2: Local Maxima, Local Minima, Saddle Points

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 Derivative

\[f_x = 24 - 6y^2\] \[f_y = -12xy - 24y^2\]

Critical Points

Set \(f_x = 0\)

\[24-6y^2=0\] \[24=6y^2\] \[4=y^2\] \[\pm2 = y\] Substitute y-values into \(f_y =0\)

\[-12xy - 24y^2=0\] \[xy+2y^2=0\] Substitute y=2 \[x(2) + 2(2)^2 = 0\]

\[2x+8=0\] \[2x=-8\] \[x=-4\]

Substitute y=-2 \[x(-2) + 2(-2)^2 = 0\]

\[-2x+8=0\] \[-2x=-8\] \[x=4\] Substitute (x,y) values for z \[ f(x,y) = 24x-6xy^2-8y^3\] Substitute (-4,2)

\[ z = 24(-4)-6(-4)(2)^2-8(2)^3\] \[z = -96+96-64\] \[z=-64\] Substitute (4,-2)

\[ z = 24(4)-6(4)(-2)^2-8(-2)^3\] \[z = 96-96+64\] \[z=64\]

Critical points are (-4,2,-64) and (4,-2,64)

Referring to the definition in page 752 of the textbook

\[f_{xx} = 0\] \[f_{yy} = -12x - 48y\] \[f_{xy} = - 12y\] Plug into formula \[ D = f_{xx}(x,y)*f_{yy}(x,y) - f_{xy}^2(x,y)\]

\[ D = 0(x,y)*(-12x - 48y(x,y) )- (- 12y)^2(x,y)\] \[ D = 0- 144y^2(x,y)\] \[D = -144y^2(x,y)\]

If we plug in y=2 or y=-2, we get D<0 concluding critical points (-4,2,-64) and (4,-2,64) are both saddle points.

Problem 3: Revenue Function

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 \(R ( x, y )\).

Revenue = units sold x price

\[R(x,y) = x(81-21x+17y) + y(40+11x-23y)\]

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

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

Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?

\[R(2.3,4.1)=-21(2.3)^2-23(4.1)^2+28(2.3)(4.1)+81(2.3)+40(4.1)\]

## [1] "$ 116.62"

Problem 4: Minimize Costs

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?

Given \(x+y=96\)

y=96-x

Substitute (96-x) into y \[C(x,y)= \frac{1}{6}x^2 + \frac{1}{6}y^2 +7x+25y+700\]

\[C(x,y)= \frac{1}{6}x^2 + \frac{1}{6}(96-x)^2 +7x+25(96-x)+700\] \[C(x,y)= \frac{1}{6}x^2 + \frac{1}{6}(9216-192x+x^2) +7x+2400-25x+700\] \[C(x,y)= \frac{1}{6}x^2 + 1536-32x+\frac{x^2}{6} +7x+2400-25x+700\] \[C(x,y)= \frac{1}{3}x^2-50x+4636\] Similar to Question 5, we need to set derivative = zero.

\[C'= \frac{2}{3}x-50\] \[0 = \frac{2}{3}x-50\] \[50 = \frac{2}{3}x\]

\[75 = x\]

Thus the number of units produced in Los Angeles to minimize the total weekly cost is 75 units. If a total of 96 units are produced each week, the remaining 21 units are produced in Denver (\(y=21\)).

Problem 5: Double Integral

Evaluate the double integral on the given region.

\[\underset{R}{\iint} (e^{8x+3y})dA;R:2\le x\le4, 2\le y\le4\]

Write your answer in exact form without decimals.

u=8x+3y, du = 8

\[\ \frac{1}{8}\int_{2}^{4} e^u du \]

\[=\left. \ \frac{e^{8x+3y}}{8}\right|_{2}^{4}\] \[=\ \frac{e^{8(4)+3y}}{8} - \frac{e^{8(2)+3y}}{8}\] \[=\ \frac{e^{32+3y}}{8} - \frac{e^{16+3y}}{8}\]

\[\ \int_{2}^{4} \frac{e^{32+3y}-e^{16+3y}}{8} dy \] Separate into two integrals

\[ \frac{1}{8}\int_{2}^{4} e^{32+3y} dy -\frac{1}{8}\int_{2}^{4} e^{16+3y} dy\]

\[=(\frac{1}{8})\left. \ \frac{e^{32+3y}}{3}\right|_{2}^{4}-(\frac{1}{8})\left. \ \frac{e^{16+3y}}{3}\right|_{2}^{4}\] \[=\ (\frac{1}{8})(\frac{e^{32+3(4)}}{3} - \frac{e^{32+3(2)}}{3})- (\frac{1}{8})(\frac{e^{16+3(4)}}{3} - \frac{e^{16+3(2)}}{3})\]

\[=\ (\frac{1}{8})(\frac{e^{32+12}}{3} - \frac{e^{32+6}}{3})- (\frac{1}{8})(\frac{e^{16+12}}{3} - \frac{e^{16+6}}{3})\] \[=\ \frac{e^{44}}{24} - \frac{e^{38}}{24}- \frac{e^{28}}{24} + \frac{e^{22}}{24}\] \[ = \frac{e^{44}-e^{38}-e^{28}+e^{22}}{24}\]