# $$\color{red}{( 5.6, 8.8 ), ( 6.3, 12.4 ), ( 7, 14.8 ), ( 7.7, 18.2 ), ( 8.4, 20.8 )}$$

mydf <- data.frame(x=c(5.6, 6.3, 7, 7.7, 8.4), y=c(8.8, 12.4, 14.8, 18.2, 20.8))
mydf
##     x    y
## 1 5.6  8.8
## 2 6.3 12.4
## 3 7.0 14.8
## 4 7.7 18.2
## 5 8.4 20.8
lmod = lm(y ~ x, data=mydf)
summary(lmod)
##
## Call:
## lm(formula = y ~ x, data = mydf)
##
## 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
coef(lmod)
## (Intercept)           x
##  -14.800000    4.257143

From the stats above.. Equation is : y = (4.26 * x) - 14.8

# 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. $$\color{red}{f(x,y) = 24x - 6xy^2 - 8y^3}$$

First derivative:

$$f_x(x, y) = 24 - 6y^2$$

$$f_y(x, y) = -12xy - 24y^2$$

For $$24-6y^2=0$$, then $$y^2 = 4$$ and $$y = \pm2$$.

For $$y=2$$ and $$-12xy - 24y^2=0$$, then $$-24x = 24\times 4$$ and $$x=-4$$.

For $$y=-2$$ and $$-12xy - 24y^2=0$$, then $$24x = 24\times 4$$ and $$x=4$$.

Replacing x and y

$$f(4,-2) = 24\times 4 - 6\times4\times (-2)^2 - 8 \times (-2)^3 = 64$$

$$f(-4,2) = 24\times (-4) - 6\times(-4)\times 2^2 - 8 \times 2^3 = -64$$

The critical points are $$(4,-2,64)$$ and $$(-4, 2, -64)$$.

Second Derivative test to determine if points are minimum, maximum or saddle.

$$f_{xx}=0$$

$$f_{yy}=-12x-48y$$

$$f_{xy}=-12y$$

$$D(x,y) = f_{xx} f_{yy}-f^2_{xy} = -(-12y)^2 = -144y^2$$.

$$D(x,y)<0$$ for all $$(x, y)$$, any critical point is a saddle point.

# 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 ).

$\begin{split} R(x,y) &= (81 - 21x + 17y)x + (40 + 11x - 23y)y \\ R(x,y) &= 81x-21x^2+17xy+40y+11xy-23y^2\\ R(x,y) &=81x+40y+28xy-21x^2-23y^2 \end{split}$

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

$$R(2.3, 4.1)=81\times 2.3+40\times 4.1+28\times 2.3\times 4.1-21\times (2.3)^2-23\times (4.1)^2 = 116.62$$

Total revenue: \$116.62.

# 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 $$\color{red}{C(x,y)=\frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700}$$, 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?

$$x+y=96$$, then $$x=96-y$$.

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

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

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

$$C_1'(y) = \frac{2}{3}y-14$$

For minimal value of cost:- $$C_1'(y)=\frac{2}{3}y-14=0$$, so $$y=21$$ or $$x=96-y=75$$.

75 units produced in Los Angeles and 21 units produced in Denver.

# Evaluate the double integral on the given region. $$\color{red}{\int\int_R e(^{8x+3y})dA ; R: 2 \leq x \leq 4}$$ and $$\color{red}{2 \leq y \leq 4}$$.

$$\int\int(e^{8x + 3y})DA: R:2<x<=4$$ and $$2<y<=4$$

$$\int_{2}^{4}\int_{2}^{4}(e^{8x + 3y})DA$$

$$\int_{2}^{4}\int_{2}^{4}(e^{8x + 3y}) \frac {3dy}{3}.dx$$

$$=\frac{1}{3}\int_{2}^{4}\bigg[(e^{8x + 3y})\bigg]_{2}^{4}.dx$$
$$=\frac{1}{3}\int_{2}^{4} (e^{8x+12} -e^{8x +6} )dx$$

$$=\frac{1}{3}\int_{2}^{4} (e^{8x+12} -e^{8x +6} ). \frac{8}{8}dx$$

$$=\frac{1}{24} \bigg[ (e^{8x+12} -e^{8x +6} ) \bigg] _{2}^{4}$$

$$=\frac{1}{24} \bigg[ e^{44} -e^{38} -(e^{28} - e^{22}) \bigg]$$

$$=\frac{1}{24} \bigg[ e^{44} -e^{38} -e^{28} + e^{22}) \bigg]$$

$$=\frac{1}{24} . 1$$