library(tidyverse)
  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)
summary(lm(y~x))
## 
## 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

\(y= 4.26x - 14.8\)

  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.

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

d_dx = D(expr(24*x - 6*x*y^2 - 8*y^3),'x')
d_dy = D(expr(24*x - 6*x*y^2 - 8*y^3),'y')
d_dx
## 24 - 6 * y^2
d_dy
## -(6 * x * (2 * y) + 8 * (3 * y^2))

Solve for d_dx = 0

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

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

\(4 = y^2\)

\(+/-2 = y\)

Solve for d_dy = 0

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

y = 2

\(0 = -24x - 96\)

\(96 = -24x\) \(x = -4\)

y = -2

\(0 = 24x - 96\)

\(96 = 24x\) \(x = 4\)

So the two critical points are (-4,2) and (4,-2).

Second Order Derivatives

d_dxx = D(D(expr(24*x - 6*x*y^2 - 8*y^3),'x'),'x')
d_dxy = D(D(expr(24*x - 6*x*y^2 - 8*y^3),'x'),'y')
d_dyy = D(D(expr(24*x - 6*x*y^2 - 8*y^3),'y'),'y')

d_dxx
## [1] 0
d_dxy
## -(6 * (2 * y))
d_dyy
## -(6 * x * 2 + 8 * (3 * (2 * y)))

(d_dxx * d_dyy) - d_dxy = (0 * -12xy - 24y^2) - (-12y)^2 = -144y^2 = +/- 576

Both of these points are saddle points.

Plugging back into the original equation, we have:

\(z = 24(-4) - 6(-4)(2)^2 - 8(2)^3 = -64\)

\(z = 24(4) - 6(4)(-2)^2 - 8(-2)^3 = 64\)

\((-4,2,-64), (4,-2,64)\)

  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).

\(R = -21x^2 + 17xy + 81x -23y^2 + 11xy + 40y\)

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

x = 2.30
y = 4.10
x*(81 - 21*x + 17*y) + y*(40 + 11*x - 23*y)
## [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) = 1/6 x 2 + 1/6 y 2 + 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?

\(C(x, y) = \frac1{6} x 2 + \frac1{6} y 2 + 7x + 25y + 700\)

\(x + y = 96\)

\(y = 96-x\)

\(C(x) = \frac1{6} x 2 + \frac1{6} (96-x) 2 + 7x + 25(96-x) + 700\)

Simplify

\(C(x)=\frac1{3}x^2−50x+4636\)

\(C'(x) = \frac2{3}x - 50\)

\(x = 75\)

\(C(x)=\frac{75}{3}^2−50(75)+4636 = 894.33\)

75 units should be manufactured in Los Angeles and 21 units should be manufactured in Denver to minimize the total weekly cost to $894.33.

  1. Evaluate the double integral on the given region.

\(\int_{2}^{4}\int_{2}^{4} e^{8x + 3y} dA ; R: (2 <= x <= 4), (2 <= y <= 4)\)

\(u = 8x - 3y\) \(du = 8\)

\(\frac1{8}\int{e^udu}\)

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

\(\int_{2}^{4} \frac{e^{32_3y}- e^{16+3y}}{8}\)

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

\(\frac1{8}\frac{e^{32+12}- e^{32+6}}{8} - \frac1{8}\frac{e^{16+12}- e^{16+6}}{8}\)

\(\frac{e^{44} - e^{38} - e^{28} + e^{22}}{24}\)

Write your answer in exact form without decimals.