Answer 1

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

##     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

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 0.0000897 ***
## ---
## 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: 0.00008971

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

The linear regression model is:

y = -14.80 + 4.26x

Answer 2

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

• Let’s find the critical points, which is when \(\nabla z = \overrightarrow{0}\).

Take the partial derivative with respect to x: \[\frac{\partial z}{\partial x} = 24 -6y^2 =0\] \[6y^2=24\] \[y^2=4\] \[y = \pm 2\]

Take the partial derivative with respect to y: \[\frac{\partial z}{\partial y} = -12xy -24y^2 =0\] \[xy + 2y^2=0\] \[x = -2y\]

If y = -2 :\(x=4\) If y = 2 : \(x=-4\)

\[x = \pm 4\]

Therefore, the critical points to this equation (x,y) is: (-4,2) and (4,-2).

To find whether or not if the critical points are minima, maxima or saddles, we need to take the partial second derivative for each critical point to find the concavity. If the concavity is positive, the critical point is a minima; if the concavity is a negative, the critical point is a positive.

• Let’s now find the critical points, which is when \(\nabla^2 z = \overrightarrow{0}\).

In respect to x: \[\frac{\partial^2 z}{\partial x^2}=0\]

In respect to y: \[\frac{\partial^2 z}{\partial y^2}=-12x + 48y^2=0\]

• Now let’s find \(f_{xy}\):

\[f_{xy} = -12y\]

In order to find whether or not these points are maxima, minima, or saddle points, we will need to use the equation below:

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

\(D = 0*(-12x-48y^2)-(-12y)^2 = -144y^2\)

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

\[D=-144(2)^2 <0 \]

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

\(D=-144(-2)^2 < 0\)

since for both points D < 0, both of our critical points are saddle points of f.

Answer 3

• Step 1 : 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(paste0("Revenue: $", round(rev_return, 2)))

## [1] "Revenue: $116.62"

Answer 4

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

\[ x + y = 96 \]

\[ y = 96 - x \]

Substitute y into the formula C(x,y):

\[C(x,y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700 = \frac{1}{6}x^2 + \frac{1}{6}(96-x)^2 + 7x + 25(96-x) + 700 = \frac{1}{3}x^2 - 50x + 4636\]

Find the critical points in function C(x):

\[C'(x) = \frac{2}{3}x - 50\] \[x = 75\] \[y = 96-x=21\]

Given that there are only one set of critical points, x=75 and y=21, these are the optimal production amounts. Therefore, 75 units should be produced in Los Angeles, and 21 units should be produced in Denver to minimize the costs.

C <- function(x){1/3*(x^2) -50*x + 4636}
plot(C(0:96), type='l')
points(75, 2761, pch=19, col="red")

Answer 5

\[\int\int_R (e^{8x + 3y}) dA \]

Above can be rewritten using double integrate as:

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

where \(dA = {\partial x}{\partial y} = dx dy\)

\[\int_{y=2}^{y=4} \int_{x=2}^{x=4} (e^{8x+3y}) dx dy = \int_2^4 \bigg [ \int_2^4 (e^{8x+3y}) dx \bigg] dy\] \[=\int_2^4 \bigg [ \frac{1}{8}\int_2^4 (e^{8x+3y}) 8dx \bigg] dy\] \[=\int_2^4 \bigg [ \frac{1}{8} (e^{8x+3y}) \bigg |_2^4 \bigg] dy\] \[=\int_2^4 \bigg [ \frac{1}{8} [ (e^{8(4)+3y}) - (e^{8(2)+3y}) ] \bigg] dy\] \[=\int_2^4 \bigg [ \frac{1}{8} [ (e^{32}e^{3y}) - (e^{16}e^{3y}) ] \bigg] dy\] \[=\int_2^4 \bigg [ \frac{1}{8} e^{3y}(e^{32} - e^{16}) \bigg] dy\] \[= \frac{1}{8} (e^{32} - e^{16}) \int_2^4 \frac{1}{3}e^{3y} 3dy\] \[= \frac{1}{8} (e^{32} - e^{16}) \frac{1}{3} \bigg [ e^{3y} \bigg |_2^4 \bigg ]\] \[= \frac{1}{8} (e^{32} - e^{16}) \frac{1}{3} \bigg [ e^{3(4)} - e^{3(2)} \bigg ]\] \[= \frac{1}{24} (e^{32} - e^{16}) (e^{12} - e^6) \] \[= \frac{1}{24} (e^{44} - e^{38} - e^{28} + e^{22}) \]

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

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"

---