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

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

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

lm_model <- lm(y ~ x)
print(lm_model)

Call:
lm(formula = y ~ x)

Coefficients:
(Intercept)            x  
    -14.800        4.257  
print(summary(lm_model))

Call:
lm(formula = y ~ x)

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
plot(x,y, main = "X vs. Y", xlab = "X", ylab = "Y")
abline(lm_model, col = 'blue', lty = 2)

The linear regression model is \(\text{Linear Regression } y = -14.80 + 4.26x\)

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.

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

Take the partial derivative with respect to x:\(\frac{\partial z}{\partial x} = 24 - 6y^2\)

Take the partial derivative with respect to y: \(\frac{\partial z}{\partial y} = -12xy - 24y^2\)

Now let’s find the critical points, which is when \(\nabla z = \vec{0}\)

wrtx

\(\frac{\partial z}{\partial x} = 24 - 6y^2 = 0\)

\(4 - y^2 = 0\)

\(y^2 = 4\)

\(y= 2 and -2\)

wrty

\(\frac{\partial z}{\partial y} = -12xy - 24y^2 = 0\)

\(xy + 2y^2 = 0\)

with y=\(-2x + 8 = 0\),\(-2x + 8 = 0\), \(x=4\)

with y=\(x(2) + 2(2)^2 = 0\), \(2x+8=0\),\(x=-4\)

Hence, the critical points to this equation is: (−4,2)and(4,−2).

\(D = f_{xx}(x,y) f_{yy}(x,y) - f_{xy}(x,y)^2 \text{ where D > 0: max or min; D < 0: saddle point}\)

\(D = 0 * (-12x - 48y^2) - (-12y)^2 = 12y^2\)

\(D = 12(2)^2 > 0 \text{ and } \frac{\partial^2 z}{\partial y^2}(-4,2) = -12(-4) - 48(2)^2 = -144\)

At critical point (x,y)=(4,-2):

\(D = 12(-2)^2 > 0 \text{ and } \frac{\partial^2 z}{\partial y^2}(4,-2) = -12(4) - 48(-2)^2 = -240\)

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)
}

rev_return <- revenue(2.30, 4.10)
print(round(rev_return, 2))
[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)=1/6x2+1/6y2+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?

Total number of units produced is 96, x from Los Angeles and y from Denver.

\(x + y = 96\) \(y = 96 - x\)

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

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

\(C(x) = \frac{1}{6} x^2 + \frac{9216 -192x + x^2}{6} + 7x + 2400 - 25x + 700\)

\(C(x) = \frac{1}{6} x^2 + 1536 - 32x + \frac{x^2}{6} + 7x + 2400 - 25x + 700\)

\(C(x) = \frac{1}{3} x^2 - 50x + 4636\)

critical points in function C(x):

\(C'(x) = \frac{2}{3} x - 50\)

\(\frac{2}{3} x - 50 = 0\)

\(x = 75\)

After, plugging in values 75+y=96

\(y=21\)

5-Evaluate the double integral on the given region.

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

Let u =8x

\(\frac{du}{dx} = 8 \text{ which translates to } du = 8 \text{ }dx\)

\(\int_{y = 2}^{y = 4} \int_{u = 16}^{u = 32} e^{ey}e^u * \frac{1}{8} du \text{ } dy = \int_{y = 2}^{y = 4} \frac{e^{ey}}{8} \int_{u = 16}^{u = 32} e^u du \text{ }dy\)

\(\int_{u = 16}^{u = 32} e^u du = e^u|_{16}^{32} = e^{32} - e^{16}\)

\(\int_{y=2}^{y=4} \frac{e^{32} - e^{16}}{8} e^{ey} \text{ }dy = \frac{e^{32} - e^{16}}{8} \int_{y=2}^{y=4} e^{ey} \text{ } dy\)

\(\frac{e^{32} - e^{16}}{8} \int_{u=2e}^{u=4e} \frac{1}{e}e^{u} \text{ } du = \frac{e^{32} - e^{16}}{8e} \int_{u=2e}^{u=4e} e^{u} \text{ } du\)

\(= (\frac{e^{31} - e^{15}}{8}) e^u|_{2e}^{4e} = (\frac{e^{31} - e^{15}}{8}) (e^{4e}-e^{2e}) + c\)

Hence, \((\frac{e^{31} - e^{15}}{8}) (e^{4e}-e^{2e}) + c\)