DATA605 Week 15

Question:

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

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

Answer:

First, let’s look at the data:

So, it looks like we know that there is a linear regression that would fit this data approximately well. My interpretation of this question is that we wish to determine the best linear regression fit for this set of data using calculus methodologies. So here we go:

\[Error = \sum {{\left(\hat{y} - y \right)}^{2}} = {\left(mx + b - y\right)}^{2}\]

\[= {m}^{2}{x}^{2} + {b}^{2} + {y}^{2} + 2mxb - 2mxy - 2by\]

\[= {m}^{2} \sum{{x}^{2}} + \sum_{x,y}{b}^{2} + \sum{{y}^{2}} + 2mb\sum{x} - 2m\sum{xy} - 2b\sum{y}\]

Now comes the calculus, we can find the minimum for each variable by taking the derivative. Note, the \({b}^{2}\) summary term is just a count of observations

\[\frac{dError}{dm} = 2m\sum{{x}^{2}} + 2b\sum{x} - 2\sum{xy} = 0\]

\[\frac{dError}{db} = 2b\sum_{x,y}{} + 2m\sum{x} - 2\sum{y} = 10b + 2m\sum{x} - 2\sum{y} = 0\]

We can now compute the individual sums for x, y, and xy:

So, our equations now look as follows:

\[\frac{dError}{dm} = 2m*249.9 + 2b*35 - 2*545.86 = 499.8m + 70b - 1091.72 = 0\]

\[\frac{dError}{db} = 10b + 2m * 35 - 2 * 75 = 10b + 70m - 150 = 0\]

We can now substitute \(\frac{dError}{db}\) into \(\frac{dError}{dm}\).

\[\frac{150 - 70m}{10} = 15 - 7m = b\]

\[499.8m + 70\left( 15 - 7m \right) - 1091.72 = 499.8m + 1050 - 490m - 1091.72 = 9.8m - 41.72 = 0 \rightarrow m = 4.26\]

using \(\frac{dError}{db}\) we can get b:

\[b = 15 - 7m = 15 - 7*4.26 = -14.82\]

So, let’s plot it.

Question:

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.

\[f\left(x,y\right) = 24x - 6x{y}^{2} - 8{y}^{3}\]

Answer:

First, we need to determine some partial and second derivatives for this function:

\[f_{x}\left(x,y\right)=24-6{y}^{2}\] \[f_{y}\left(x,y\right)=-12xy-24{y}^{2}\] \[f_{xx}\left(x,y\right)=0\]

\[f_{yy}\left(x,y\right)=-12x-48y\] \[f_{xy}\left(x,y\right)=-12y\]

Okay, so now we can find the critical points by taking the partial derivatives and setting them to zero:

\[f_{x}\left(x,y\right)=24-6{y}^{2} = 0 \rightarrow y = \sqrt{\frac{24}{6}} = \pm2\] For a value of \(y = 2\):

\[f_{y}\left(x,y\right)=-12xy-24{y}^{2} = 0 = -24x-96 \rightarrow y = -4\]

So, we only have two critical points which are \(\left( -4, 2\right), \left( 4, -2\right)\).

now, let’s find the results of the second derivative test for each point:

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

So, we can see here that \(D\left(x,y\right) < 0\) for all \(y \ne 0\). Therefore, we know that both critical points are saddle points, but let’s graph them for fun.

Question:

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 \(R ( x, y )\).

Answer:

\[hb = 81-21x+17y, nb = 40+11x-23y\]

The revenue would be the number of units multiplied by the sale price for each brand:

\[R\left(x,y\right) = x * \left( 81-21x+17y \right) + y*\left( 40+11x-23y \right)\] \[= 81x -21{x}^{2}+17xy+40y+11xy-23{y}^{2} = -21{x}^{2} - 23{y}^{2} + 81x + 40y + 28xy\]

Question:

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

Answer:

The revenue is:

## [1] 116.62

Question:

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?

Answer:

Units function:

\[U\left( x,y \right) = x + y = 96\]

We can substitute the units function into the cost function to get a formula with one variable:

\[\frac{1}{6}{x}^{2}+\frac{1}{6}{\left( 96 - x\right)}^{2}+7x+25\left( 96 - x\right) +700\]

\[= \frac{{x}^{2}}{6}+1536 - 32x + \frac{{x}^{2}}{6}+ 7x + 2400 - 25x + 700 = \frac{{x}^{2}}{3}-50x+4636\]

So, we can take the derivative and evaluate at zero:

\[\frac{d}{dx} = \frac{2}{3}x-50 =0 \rightarrow x = 75\]

We can use the units function to determine y:

\[96 - 75 = 21\]

Therefore, to maximize cost, 75 units should be produced in Los Angeles and 21 units should be produced in Denver. Let’s graph it just for fun:

Question:

5. Evaluate the double integral on the given region.

\[\iint _{ R }^{ }{ \left( {e}^{8x+3y} \right)dA;\quad R:2\le x\le4\quad and \quad 2\le y\le 4 } \]

Write your answer in exact form without decimals.

Answer:

So, we integrate over one variable, then the other:

\[\int_{2}^{4} {\int_{2}^{4} {\left( {e}^{8x+3y} dx \right)dy}}\]

The first part of this requires substitution which is \(u = 8x + 3y\) so \(du = 8 dx, \frac{du}{8}=dx\)

\[\int_{2}^{4} {\int_{2}^{4} {\left(\frac{{e}^{u}du}{8} \right)dy}} = \int{\left[ \frac{{e}^{8x+3y}}{8} \right]_{2}^{4}dy}\]

\[= \int{\left(\frac{{e}^{32+3y}}{8} - \frac{{e}^{16+3y}}{8}\right)dy} = \frac{\left( {e}^{16}-1 \right)\left( {e}^{16+3y} \right)}{8}\]

\[= \frac{\left( {e}^{16}-1 \right)}{8}\int_{2}^{4}{{e}^{16+3y}dy}\]

Now, we do u subsitution again \(u = 3y + 16, dy = \frac{du}{3}\):

\[= \frac{{e}^{16} - 1}{24} \int_{2}^{4}{{e}^{u}du}\]

\[= \frac{{e}^{16} - 1}{24} \left( {e}^{28} - {e}^{22} \right)\]