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)

lin_reg = lm(y ~ x)
summary(lin_reg)
## 
## 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)
abline(lin_reg)

Equation of the regression line:

\[B_0 = -14.800 \\ B_1 = 4.257 \\ y = B_0 + B_1x \\ y = -14.800 + 4.2571x\]

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

Find the partial derivatives of f:

\[f_x(x, y) = 24 + 6y^2 ; \\ f_y(x,y) = -12x - 24y^2\]

Set each equal to 0 and solve for x and y:

\[f_x(x, y) = 0 => y = +-2 ; \\ f_y(x,y) = 0 => x = -+4\]

Therefore, we have two critical points: (-2, 4), (2, -4). To determine if it corresponds to a relative maximum, minimum, or saddle point, we will look at the second partial derivative of f.

Let z = f(x, y) be differentiable on an open set containing P = (x_0 , y_0), and let

\[D = f_{xx}(x_0, y_0)f_{yy}(x_0, y_0) − f_{xy}^2(x_0, y_0)\]

  1. If D < 0 and \[f_{xx}(x_0,y_0) > 0\], then P is a relative minimum of f

  2. If D > 0 and \[f_{xx}(x_0,y_0) < 0\], then P is a relative maximum of f

  3. If D<0, then P is a saddle point of f

  4. If D = 0, the test is inconclusive

\[f_{xx} = 24 \\ f_{yy} = 48y \\ f_{xy} = 12y\]

For (-2, 4):

\[f_{xx} = 24 \\ f_{yy} = 192 \\ f_{xy} = 48\]

Therefore, \[D(x,y) = 4560\] and (-4, 2) is a relative maximum of f.

For (2, -4):

\[f_{xx} = 24 \\ f_{yy} = -192 \\ f_{xy} = -48\]

Therefore, \[D(x,y) = -4560\] and (-4, 2) is a relative minimum of f.

  1. 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 ).

Revenue for House brand: \[R(x)=x∗(81−21x+17y)\]

Revenue for name brand: \[R(y)=y∗(40+11x−23y)\]

Total Revenue: \[R(x,y)=x∗(81−21x+17y)+y∗(40+11x−23y)\]

\[R(x,y)=81x−21x^2+17xy+40y+11xy−23y^2\]

\[R(x,y)= - 21x^2 - 23y^2 + 28xy + 81x + 40y \]

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

\[R(2.3,4.1)= - 21x^2 - 23y^2 + 28xy + 81x + 40y\]

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

revenue(2.3, 4.1)
## [1] 116.62
  1. 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 \[(x,y)= \dfrac{1}{6}x^2 + \dfrac{1}{6}y^2 +7x+25y+700\], where x is the number of 66 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?

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

\[ = \dfrac{1}{6}(96-y)^2 + \dfrac{1}{6}y^2 +7(96-y)+25y+700\]

\[ = \dfrac{1}{6}(y^2-192y+9216) + \dfrac{1}{6}y^2 +672 -7y+25y+700\]

\[ = \dfrac{1}{6}y^2-32+1536 + \dfrac{1}{6}y^2 +672 -7y+25y+700\]

\[ C(x,y) = \dfrac{1}{3}y^2-14y + 2908\]

\[ C'(x,y) = \dfrac{2}{3}y-14\]

At C’ = 0; y = 21, x = 96 - y = 75

  1. Evaluate the double integral on the given region.

\[\int\int_R (e^{8x+3y})dA; \ R: 2≤x≤4 \ and \ 2≤y≤4\]

Write your answer in exact form without decimals.

\[\int_2^4\int_2^4 (e^{8x} + e^{3y})dxdy\]

\[\int_2^4 e^{8x}dx * \int_2^4 e^{3y}dy\]

\[\dfrac{1}{8}e^{8x}dx|_2^4 * \dfrac{1}{3}e^{3y}dy|_2^4\]

\[\dfrac{1}{24}e^{8x}|_2^4 * e^{3y}|_2^4\]

\[\dfrac{1}{24}(e^{32}-e^{16})(e^{12}-e^6)\]

1/24*((exp(32)+exp(16))*(exp(12)-exp(6)))
## [1] 5.341561e+17