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 )

Ans:

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-05

Equation of the regression line for the given points is:

\(y = -14.80 + 4.26x\)

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 - 6xy² - 8y³

Ans:

\[ \begin{multline*} \begin{split} f_x &= 24 - 6y^2 \\ f_y &= -12xy - 24y^2 \\ \end{split} \end{multline*} \]

For finding critical points, we set the partial derivatives to 0s.

\[ \begin{multline*} \begin{split} 24 - 6y^2 = 0 \\ -12xy - 24y^2 = 0 \\ \end{split} \end{multline*} \]

results in \(\begin{Bmatrix} y = \pm 2 \\ \begin{Bmatrix} x = -4 \hspace{8pt} if \hspace{8pt} y = 2 \\ x = 4 \hspace{8pt} if \hspace{8pt} y = -2 \\\\ \end{Bmatrix}\end{Bmatrix}\)

The critical points are (-4,2) and (4, -2)

Taking second derivatives of f gives us:

\[ \begin{multline*} \begin{split} f_{xx} &= 0\\ f_{yy} &= -12x-48y \\ f_{xy} &= -12y \\ \end{split} \end{multline*} \]

Now using 2nd derivative test, we get

D = \(f_{xx}f_{yy} - f_{xy}^2 = -144y^2\)

At c.p. (-4,2), D = -576 < 0 [saddle point]

At c.p. (4,-2), D = -576 < 0 [saddle point]

So we found 2 saddle points, local maxima or minima.

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.

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?

Ans:

The total weekly cost is given by the following function:

C(x,y) = \(\frac{1}{6}x^2+\frac{1}{6}y^2+7x+25y+700\)

where xx is the number of units produced in Los Angeles and yy is the number of units produced in Denver.

x + y = 96

x = 96 - y

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

\(= \frac{1}{3}y^2-14y+2908\)

C’(x,y) = \(\frac{2}{3}y - 14\)

When C’(x,y) = 0

\(\frac{2}{3}y- 14 = 0\)

\(y = 21\)

x = 75

Therefore, the company needs to produce 75 units in Los Angeles and 21 units in Denver to minimize the total weekly cost.

5.

Evaluate the double integral on the given region.

Write your answer in exact form without decimals.

Ans:

Let u = 8x+3y

\(\frac{\mathrm{du} }{\mathrm{d} y}=3\)

\(dy = \frac{1}{3}du\)

\[ \begin{multline*} \begin{split} \int_{2}^{4}\int_{2}^{4}e^{8x+3y}dxdy &= \int_{2}^{4}\frac{1}{3}\int_{2}^{4}e^udu dx \\ &= \int_{2}^{4}\frac{1}{3}\int_{2}^{4}e^{8x+3y}dydx \\ &= \int_{2}^{4}\frac{1}{3}(e^{8x+12}-e^{8x+6})dx \\ &= \frac{1}{3}\times\frac{1}{8}(e^{8(4)+12} -e^{8(2)+12}) - \frac{1}{3}\times\frac{1}{8}(e^{8(4)+6} -e^{8(2)+6}) \\ &= \frac{1}{24}(e^{32+12}-e^{16+12}-e^{32+6}+e^{16+6}) \\ &= \frac{1}{24}(e^{44}-e^{28}-e^{38}+e^{22}) \\ \end{split} \end{multline*} \]