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

independent <- c(5.6, 6.3, 7, 7.7, 8.4)
dependent <- c(8.8, 12.4, 14.8, 18.2, 20.8)

model <- lm(dependent ~ independent)
summary(model)
## 
## Call:
## lm(formula = dependent ~ independent)
## 
## 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 ***
## independent   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

Solution:

\(y=-14.80 + 4.26x\)

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.

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

f=expression(24*x - 6*x*y^2 - 8*y^3)

fx <- D(f,'x')
Simplify(fx)
## 24 - 6 * y^2
fy <- D(f,'y')
Simplify(fy)
## -(y * (12 * x + 24 * y))

Solve for y

\(0 = 24 - 6y^2\)

\(6y^2 = 24\)

\(6y^2 = 24\)

\(y^2 = 4\)

\(y \pm 2\)

Solve for x

\(0 = -12y (x+2y)\)

\(0 = -12yx -24y^2\)

\(12yx = -24y^2\)

\(x = \frac{-24y^2}{12y}\)

\(x = -2y\)

\(x = -2(\pm2)\)

\(x \pm 4\)

fxx <- D(D(f,'x'),'x')
Simplify(fxx)
## [1] 0
fxy <- D(D(f,'x'),'y')
Simplify(fxy)
## -(12 * y)
fyy <- D(D(f,'y'),'y')
Simplify(fyy)
## -(12 * x + 48 * y)

\(0 * -12(x+4y) - (-12y)^2\)

\(-144y^2\)

Evaluated at with 2 and -2 (x is ignored due to no term) the result is: -576.

Since, -576 is less than zero, (-4, 2) and (4, -2) are saddle points.

There is no local maxima or local minima.

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 ).

house <- expression(81-21*x+17*y)
name <- expression(40 + 11*x - 23*y)

house_and_name <- expression(x*(81-21*x+17*y) + y*(40 + 11*x - 23*y))

Simplify(house_and_name)
## expression(x * (17 * y + 81 - 21 * x) + y * (11 * x + 40 - 23 * 
##     y))

Expanding out… \(17xy + 81x - 21x^2 + 11xy + 40y - 23y^2\)

Revenue function is: \(28xy + 81x - 21x^2 + 40y - 23y^2\)

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

x*(81 - 21*x + 17*y) + y*(40 + 11*x - 23*y)
## [1] 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 x2 + 1/6 y2 + 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?

Cost function is:

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

Required output per week can be defined as: \(x + y = 96\)

Which we can rewrite as: \(x = 96 - y\)

Let’s substitute the expression above for x:

\(\frac{1}{6} (96 - y)^2 + \frac{1}{6} y^2 + 7(96 - y) + 25y + 700\)

Which simplifies to:

\(1536 - 32y + \frac{1}{6}y^2 + \frac{1}{6} y^2 + 672 - 7y + 25y + 700\)

\(2908 - \frac{1}{3}y^2 - 14y\)

Let’s take the first derivative to find minimum cost…

f <- expression(2908 - (1/3)*y^2-14*y)
fx <- D(f,'y')
Simplify(fx)
## -(0.666666666666667 * y + 14)

\(0 = -\frac{2}{3}y - 14\)

Solve for y:

\(y = 21\)

Therefore, if 21 units should be produced in Denver..

\(x + 21 = 96\)

Then, the amount of units produced in Los Angeles should be:

\(x = 75\)

5) Evaluate the double integral on the given region. Write your answer in exact form without decimals.

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

Apply u-substitution

u = \(8x+3y\)

Take derivative with respect to x:

\(8+0\)

\(\frac{du}{dx} = 8+0\)

\(dx = \frac{1}{8}du\)

Now plug in 2 and 4 to u for adjusted boundaries:

\(8(2) + 3y = 16 + 3y\)

\(8(4) + 3y = 32 + 3y\)

The updated double integral becomes:

\(\int_{16 + 3y}^{32 + 3y} e^u \frac{1}{8} \,du\)

Move the constant to the front:

\(\frac{1}{8} \int_{16 + 3y}^{32 + 3y} e^u \,du\)

\(\frac{1}{8} (e^{32 + 3y} - e^{16 + 3y}) \,du\)

\(\int_{2}^{4} \frac{1}{8} (e^{32 + 3y} - e^{16 + 3y}) \,dy\)

Apply sum rule

\(\frac{1}{8} \int_{2}^{4} e^{32 + 3y} - \int_{2}^{4}e^{16 + 3y} \,dy\)

Apply u-substitution

u = \(32 + 3y\)

Take derivative with respect to y:

\(0+3\)

\(\frac{du}{dy} = 0+3\)

\(dy = \frac{1}{3}du\)

Now plug in 2 and 4 to u for adjusted boundaries:

\(32 + 3(2) = 38\)

\(32 + 3(4) = 44\)

The updated double integral becomes:

\(\int_{38}^{44} e^u \frac{1}{3} \,du\)

\(\frac{1}{3} (e^{44} - e^{38})\)

Apply u-substitution

u = \(16 + 3y\)

Take derivative with respect to y:

\(0+3\)

\(\frac{du}{dy} = 0+3\)

\(dy = \frac{1}{3}du\)

Now plug in 2 and 4 to u for adjusted boundaries:

\(16 + 3(2) = 22\)

\(16 + 3(4) = 28\)

The updated double integral becomes:

\(\int_{22}^{28} e^u \frac{1}{3} \,du\)

\(\frac{1}{3} (e^{28} - e^{22})\)

Now combine:

\((e^{44} - e^{38}) - (e^{28} - e^{22})\)

\(e^{44} - e^{38} - e^{28} + e^{22}\)

Insert the constants:

\(\frac{1}{8}(\frac{1}{3}(e^{44} - e^{38} - e^{28} + e^{22}))\)