Problem 1

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 )

A <- c(5.6, 6.3, 7, 7.7, 8.4)
x <- c(8.8,12.4,14.8,18.2,20.8)
mtx <- matrix(cbind(A,x), ncol=2)

# Ax = b where x = intercept, b

model <- lm(A~x)
model
## 
## Call:
## lm(formula = A ~ x)
## 
## Coefficients:
## (Intercept)            x  
##      3.4890       0.2341

y = 3.4890 + .2341x # Problem 2 \[ f(x,y) = 24x-6xy^2-8y^3 \implies \\ \frac{df}{dx}f(x,y) = 24-6y^2 = \frac{df}{dy}f(x,y) = -12xy-24y^2 = 0 \implies \\ 24-6y^2 = 12xy-24y^2 \implies 24 = 12xy-18y^2 \] Which means that \[ -6y^2+24 = 0, \\ -12xy-24y^2 = 0 \\ 1-2x = 24y^2 \\ x = 2y^2 \implies \\ 3x = 24 \implies x =8, y=\pm2 \]

Critical points at \[ p_0 = (0,0)\\ p_1 = (8,-2)\\ p_2 = (8, 2)\\ \] The Hessian is \[ D^2f(x,y) = \begin{bmatrix} 0 & -12y \\ -12y & 0 \end{bmatrix} \] p0 is a point of interest, but it’s status remains unknown from the second derivative test as the D(Hessian) = 0. P1 is a a is a saddle point. P2 is a local minimum.

Problem 3

Let the house brand be represented by by x and the name brand by y. We can sell \(81-21x+17y\) units of house and \(40+11x-23y\) units of name brand. We can then compute the revenue as a sum of products. \[ p_1(x,y) = 81-21x+17y \\ p_2(x,y) = 40+11x-23y \]

\[R(x,y) = p_1(x,y)\cdot x + p_2(x,y)\cdot y\]

b

library(matlib)
p1 = 2.30 #p1(x1,x2)
p2 = 4.10 #p2(x1,x2)
A <- matrix(c(-21, 17, 11, -23), nrow=2)
b <- c(-81,-40)
showEqn(A,b)
## -21*x1 + 11*x2  =  -81 
##  17*x1 - 23*x2  =  -40
solve(A,b)
## [1] 7.780405 7.489865
x = 7.78
y = 7.489
R = x*p1 + y *p2
R
## [1] 48.5989

Problem 4

A company has two plants where x is the number of widgets by in plant A and y is the number of widgets made in plant B. \[ \text{cost} = f(x,y) = \frac{1}{6}x^2+\frac{1}{6}y^2+7x+25y+700 \\ \text{number of units} = g(x,y) = x + y = 96 \]

\[ \nabla f = (\frac{x}{3}+7, \frac{y+75}{3}) \\ \nabla g = (1,1) \\ \nabla f = \lambda \nabla g \implies \\ \frac{x}{3}+7 = \lambda \\ \frac{y+75}{3} = \lambda \implies \\ \frac{y+75}{3} = \frac{x}{3}+7 \implies \\ y + 75 = x + 21 \\ \text{recall that } x+ y = 96 \implies y = 96-x \implies \\ 96-x + 75 = x + 21 \implies 2x = 150 \implies x = 50 \implies \\ y = 96 - 50 = 46 \]

Problem 5

$$

2^4 {2}^{4} (e^{8x+3y})dxdy \ _2^4 e^{3y+16} dy \ = - \ =

$$