Problem 1

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

To solve this we will plug the points into a linear model function.

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

m <- lm(y~x)
summary(m)
## 
## 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

The function we get is: \[y = 4.26x - 14.8\]

Problem 2

Find all local maxima, local minima, and saddle points for the function given below \[f ( x, y ) = 24x - 6xy^2 - 8y^3\]

First lets solve all the derivatives

\[f_x=24 - 6y^2; f_y = -12xy - 24y^2\] \[f_{xx} = 0; f_{yy} = -12x - 48y; f_{xy} = -12y\]

Now set the first derivatives equal to 0

\[6y^2 = 24\] \[y = 2, -2\] \[12xy = -24y^2\] \[x = -2y\] \[x = 4, -4\]

So the two points are (2, -4) and (-2, 4). Now we need to check for which are min and max.

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

Since D < 0 in both points we are dealing with 2 saddle points.

Problem 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.

Find the revenue function R(x, y)

\[R(x,y) = [HousePrice]*[HouseUnitsSold] + [NamePrice]*[NameUnitsSold]\] \[R(x,y) = x*(81-21x+17y)+y*(40+11x-23y)\]

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

val <- x*(81-21*x+17*y) + y*(40+11*x-23*y)
val
## [1] 116.62

Problem 4

The first step is to represent C(x,y) in terms of one variable: \[y = 96 - x\] Plug this in to C(x, y) \[C(x,y) = \frac{1}{6}x^2+\frac{1}{6}(96-x)^2+7x+25(96-x)+700\] \[C(x,y) = \frac{1}{3}x^2-32x+1,536+7x+2400-25x+700\] \[C(x,y) = \frac{1}{3}x^2-50x+4,636\] Now take the derivative and set equal to zero \[\frac{dC}{dx} = \frac{2}{3}x-50\] \[\frac{2}{3}x = 50\] \[x = 75\]

To minimize the cost we need to produce 75 units in Los Angeles and 21 units in Denver

Problem 5

\[ \int_2^4\int_2^4e^{8x}e^{3y}dxdy \] \[ \int_2^4 e^{8x} *\frac{1}{3}(e^{3*4}-e^{3*2}) \] \[ \frac{1}{8}(e^{32} - e^{16}) *\frac{1}{3}(e^{12}-e^{6}) \] \[ \frac{1}{24}(e^{32+12} - e^{32+6} - e^{16+12} + e^{16+6}) \] \[ \frac{1}{24}(e^{44} - e^{38} - e^{28} + e^{22}) \]

(1/24)*(exp(44) - exp(38) - exp(28) + exp(22))
## [1] 5.341559e+17