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

x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
b0 <- round(summary(lm(y ~ x))$coefficients[1], 2)
b1 <- round(summary(lm(y ~ x))$coefficients[2], 2)

The equation of the regression line is $y=-14.8+4.26x$

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

1st order partial derivatives with respect to $x$: $$f_{x}(x,y)=24-6y^2=0$$ $$y=(2, -2)$$ 1st order partial derivatives with respect to $y$: $$f_{y}(x,y)=-12xy-24y^2$$ $$when \enspace y=2, \enspace x=-4$$ $$when \enspace y=-2, \enspace x=4$$ Therefore, the critical points of this equation are $(−4,2)$ and $(4,−2)$
2nd order partial derivatives with respect to $x$: $$f_{xx}(x,y)=0$$ 2nd order partial derivatives with respect to $y$: $$f_{yy}(x,y)=-12x-48y$$
$$\therefore f_{xy}=-12y$$
Second Derivative Test: $$D=f_{xx}(x,y)f_{yy}(x,y)-f_{xy}(x,y)^2$$
Critical point (4, -2): $$D=0*-12x-48y-(-12y)^2$$ $$D=0*-12(4)-48(-2)-(-12(-2))^2=-576$$
Critical point (-4, 2): $$D=0*-12x-48y-(-12y)^2$$ $$D=0*-12(-4)-48(2)-(-12(2))^2=-576$$
Since $D<0$, critical points $(−4,2)$ and $(4,−2)$ are saddle points.

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 $R(x,y)$.

$$\text{Revenue = (Units Sold) x (Sales Price)}$$ $$R(x,y)=(x*(81-21x+17y))+(y*(40+11x-23y))$$ $$R(x,y)=81x-21x^2+17xy+40y+11xy-23y^2$$ $$R(x,y)=-21x^2-23y^2+81x+40y+28xy$$

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

revenue <- function(x, y){
  return(-21*(x^2) - 23*(y^2) + 81*x + 40*y + 28*x*y)
}

print(paste0("The revenue if she sells the 'house' brand for $2.30 and the 'name' brand for $4.10 is $", revenue(2.3, 4.1)))

## [1] "The revenue if she sells the 'house' brand for $2.30 and the 'name' brand for $4.10 is $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 $C(x,y)=\frac{x^2}{6}+\frac{y^2}{6}+7x+25y+700$, 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?

Let us find units produced in Los Angeles by finding the first derivative of the equation with respect to $x$, $$\text{Given, } \enspace x+y=96$$ $$\therefore y=96-x$$ $$\text{Substitute y in to the equation, } \enspace C(x,y)=\frac{x^2}{6}+\frac{(96-x)^2}{6}+7x+25(96-x)+700$$ $$\therefore C(x)=\frac{x^2}{3}-50x+4636$$ 1st order derivative to get the critical points: $$C'(x)=\frac{2x}{3}-50=0$$ $$\text{Solving for x, }x=75$$ Let us now find units produced in Denver, $$75+y=96$$ $$y=21$$ Therefore, 75 units should be produced in Los Angeles and 21 units should be produced 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. $$\iint_{R}(e^{8x+3y})dA; \quad R:2\le x\le4\enspace and\enspace 2 \le y\le4$$ $$\int_{2}^{4} \int_{2}^{4} e^{8x+3y} dxdy$$ $$\int_{2}^{4} \frac{e^{3y+32}}{8}-\frac{e^{3y+16}}{8}dy$$ $$\int_{2}^{4} \frac{1}{8}(e^{16}-1)(e^{3y+16})dy$$ $$\frac{e^{44}-e^{28}}{24}-\frac{e^{38}-e^{22}}{24}$$ $$\frac{1}{24}(e^{22}-e^{28}-e^{38}+e^{44})$$

q5 <- 1/24*(exp(22)-exp(28)-exp(38)+exp(44))
print(paste0("Exact form without decimals: ", format(q5, scientific = F)))

## [1] "Exact form without decimals: 534155947497085056"