Homework Week 15

Bonnie Cooper

Homework Week 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 )
lm_dat <- lm( y ~ x )
plot( x,y ); abline( lm_dat$coefficients, col ='blue' )

paste( 'The equation for the lm best-fit line: y= ', round(summary( lm_dat )$coefficients[2,1],2), '*x +', summary( lm_dat )$coefficients[1,1])

## [1] "The equation for the lm best-fit line: y=  4.26 *x + -14.8"

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. \[f(x,y) = 24x-6xy^2 -8y^3\]

Find the first partial derivatives $f_x$ and $f_y$ \[f_x(x,y) = 24-6y^2\] \[f_y(x,y) = 12xy - 24y^2\] The critical points satisfy the fist partial derivatives set equal to zero: \[24-6y^2=0 \rightarrow\] \[24=6y^2 \rightarrow\] \[y = \sqrt{\frac{24}{6}} = \pm2\] \[12xy-24y^2=0\] solving for $y=2$: \[12x(2)-24(2)^2 = 24x - 96 = x - 4 \rightarrow x = 4\] solving for $y=-2$: \[12x(-2)-24(-2)^2 = -24x - 96 = -x - 4 \rightarrow x = -4\] Therefore, the critical points are (2,4) & (2,-4)

Find the second order partial derivatives $f_{xx}(x,y)$, $f_{yy}(x,y)$ & $f_{xy}(x,y)$ \[f_{xx}(x,y) = 0\] \[f_{yy}(x,y) = 12x-48y\] \[f_{xy}(x,y) = -12y\]

Now to use the equation $D = f_{xx}(a,b)f_{yy}(a,b) - f_{xy}^2(a,b)$ for each critical point $(a,b)$:

For point (2,4): $D = (0)(12*2-48*4) - (-12*4)^2 \rightarrow$ negative and $f_{xx}(2,4)=0$ therefore this is a saddle point
For point (2,-4): $D = (0)(12*2-48*4) - (-12*(-4))^2 \rightarrow$ negative and $f_{xx}(2,-4)=0$ therefore this is a saddle point

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 $R(x,y)$.

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

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

x <- 2.3
y <- 4.1
R <- x*( 81-21*x+17*y ) + y*( 40+11*x-23*y )
paste0( 'Revenue = $', round( R, 2 ) )

## [1] "Revenue = $116.62"

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?

It is given that: \[\mbox{Total units produced} = x + y = 96\] Can solve for x, where $x= 96-y$ and plug this expression into the Cost equation: \[C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700 \rightarrow\] \[= \frac{1}{6}(96-y)^2 + \frac{1}{6}y^2 + 7(96-y) + 25y + 700 \rightarrow\] \[= \frac{1}{3}y^2 - 14y + 2908\] Taking the 1st derivative: \[C'(y) = \frac{2}{3}y -14 \] To find the critical point: \[\frac{2}{3}y -14 = 0 \rightarrow y = 21\] Substituting back into the 1st equation: \[x + 21 = 96 \rightarrow x = 75\] To minimize cost, the LA location should produce 75 units while the Denver location should produce 21.

5)

Evaluate the double integral on the given region.

\[\int \int (e^{8x+3y})dA; R: 2 \leq x \leq 4 \mbox{ and } 2 \le y \le 4\] Write your answer in exact form without decimals.

First find the intgral with respect to x: \[\int_2^4 e^{8x+3y}dx = \frac{1}{8}(e^{3y+32}-e^{3y+16})\] Now integrate with respect to y: \[\int_2^4 \frac{1}{8}(e^{3y+32}-e^{3y+16}) dy = \frac{e^{44} -e^{38} -e^{28} + e^{22}}{24}\]