Data 605 Assignment-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 )

df <- data.frame(rbind(c( 5.6, 8.8 ), c( 6.3, 12.4 ), c( 7, 14.8 ), c( 7.7, 18.2 ), c( 8.4, 20.8 )))

lm_model <- lm(df$X2 ~ df$X1, df)

plot(df$X2 ~ df$X1, data=df)
abline(lm_model)

summary(lm_model)

## 
## Call:
## lm(formula = df$X2 ~ df$X1, data = df)
## 
## 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 ***
## df$X1         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

From the intercepts given in the summary, we can write the equation:

$y = -14.8 + 4.2571 * x$

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.

$f(x,y) = 24x - 6xy^2 -8y^3$

To get the local maxima / minima or relative extrema, we will calculate the first partial derivatives of the function on both x and y.

These derivatives will be zero at the relative extrema.

$f_{x}(x,y) = 24 - 6y^2$

$f_{y}(x,y) = -12xy - 24y^2$

At relative extrema, $f_{x}(x,y) = 0$ and $f_{y}(x,y) = 0$

$f_{x}(x,y) = 24 - 6y^2 = 0$

$y = +2, -2$

$f_{y}(x,y) = -12xy - 24y^2 = 0$

$x = -2y$

Now substituting the 2 values of y from above

At $y = +2, x = -2y = -4$

At $y = -2, x = -2y = +4$

Hence the 2 points are (-4,2) and (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).

$R(x,y) = x(81 - 21x + 17y) + y(40 + 11x - 23y)$

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

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

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

In this case, $x=2.3$, and $y=4.1$

Substituting the values of x and y in the above function:

Revenue for this case = $-21 (2.3)^2 - 23(4.1)^2 + 28(2.3)(4.1) + 81(2.3) + 40(4.1)$

$= -111.09 + 386.63 + 264.04 + 186.3 + 164$

$= 116.62$

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 $C(x,y) = (1/6) x^2 + (1/6) y^2 + 7x + 25y + 700$ , 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?

$x + y = 96$

$y = 96 - x$

Replacing y in the function C(x,y)

$C(x,96-x) = (1/6) x^2 + (1/6) (96-x)^2 + 7x + 25(96-x) + 700$

$C(x,96-x) = (1/6) x^2 + (1/6) (9216 + x^2 -192x) + 7x - 25x + 2400 + 700$

$C(x,96-x) = (1/3) x^2 + 1536 - 32x - 18x + 3100$

$C(x,96-x) = (1/3) x^2 -50x + 4636$

The above is the cost function. For C to be minimum, $C^{'} = 0$

$C^{'} = (2/3) x - 50 = 0$

$x = 75$

Substituting x in the equation : x + y = 96, $y = 21$

Hence 75 units produced in Los Angeles and 21 in Denver - to have the minimum cost

5. Evaluate the double integral on the given region.

$\int \int_R (e^{8x+3y})$ dA $; R:2 \le x \le 4$ and $2 \le y \le 4$

Write your answer in exact form without decimals.

We know that $\int e^{nx}$dx = $e^{nx}/n$

Using this, we write below:

$\int_2^4 (\int_2^4 (e^{8x+3y}) dx) dy$

$= \int_2^4 (\int_2^4 (e^{8x} e^{3y}) dx) dy$

$= \int_2^4 e^{3y} (\int_2^4 (e^{8x} ) dx) dy$

$= \int_2^4 e^{3y} \left. ((1/8)e^{8x}) \right|_{2}^{4} dy$

$= \int_2^4 (e^{3y}/8) (e^{32} - e^{16}) dy$

$= ((e^{32} - e^{16})/8) \int_2^4 e^{3y} dy$

$= ((e^{32} - e^{16})/8) \left. ((1/3)e^{3y}) \right|_{2}^{4}$

$= (1/24) (e^{32} - e^{16}) (e^{12} - e^{6})$

Data 605 Assignment-15

Deepak Mongia

May 17, 2019

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 )

\(y = -14.8 + 4.2571 * x\)

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.

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

To get the local maxima / minima or relative extrema, we will calculate the first partial derivatives of the function on both x and y.

These derivatives will be zero at the relative extrema.

\(f_{x}(x,y) = 24 - 6y^2\)

\(f_{y}(x,y) = -12xy - 24y^2\)

At relative extrema, \(f_{x}(x,y) = 0\) and \(f_{y}(x,y) = 0\)

\(f_{x}(x,y) = 24 - 6y^2 = 0\)

\(y = +2, -2\)

\(f_{y}(x,y) = -12xy - 24y^2 = 0\)

\(x = -2y\)

Now substituting the 2 values of y from above

At \(y = +2, x = -2y = -4\)

At \(y = -2, x = -2y = +4\)

Hence the 2 points are (-4,2) and (4,-2)

Step 1. Find the revenue function R(x,y).

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

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

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

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

In this case, \(x=2.3\), and \(y=4.1\)

Substituting the values of x and y in the above function:

Revenue for this case = \(-21 (2.3)^2 - 23(4.1)^2 + 28(2.3)(4.1) + 81(2.3) + 40(4.1)\)

\(= -111.09 + 386.63 + 264.04 + 186.3 + 164\)

\(= 116.62\)

\(x + y = 96\)

\(y = 96 - x\)

Replacing y in the function C(x,y)

\(C(x,96-x) = (1/6) x^2 + (1/6) (96-x)^2 + 7x + 25(96-x) + 700\)

\(C(x,96-x) = (1/6) x^2 + (1/6) (9216 + x^2 -192x) + 7x - 25x + 2400 + 700\)

\(C(x,96-x) = (1/3) x^2 + 1536 - 32x - 18x + 3100\)

\(C(x,96-x) = (1/3) x^2 -50x + 4636\)

The above is the cost function. For C to be minimum, \(C^{'} = 0\)

\(C^{'} = (2/3) x - 50 = 0\)

\(x = 75\)

Substituting x in the equation : x + y = 96, \(y = 21\)

Hence 75 units produced in Los Angeles and 21 in Denver - to have the minimum cost

5. Evaluate the double integral on the given region.

\(\int \int_R (e^{8x+3y})\) dA \(; R:2 \le x \le 4\) and \(2 \le y \le 4\)

Write your answer in exact form without decimals.

We know that \(\int e^{nx}\)dx = \(e^{nx}/n\)

Using this, we write below:

\(\int_2^4 (\int_2^4 (e^{8x+3y}) dx) dy\)

\(= \int_2^4 (\int_2^4 (e^{8x} e^{3y}) dx) dy\)

\(= \int_2^4 e^{3y} (\int_2^4 (e^{8x} ) dx) dy\)

\(= \int_2^4 e^{3y} \left. ((1/8)e^{8x}) \right|_{2}^{4} dy\)

\(= \int_2^4 (e^{3y}/8) (e^{32} - e^{16}) dy\)

\(= ((e^{32} - e^{16})/8) \int_2^4 e^{3y} dy\)

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

\(= (1/24) (e^{32} - e^{16}) (e^{12} - e^{6})\)