Question 1
##
## Call:
## lm(formula = y ~ x, data = d1)
##
## 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
Equation: $y = -14.8 + 4.46x $
Question 2
## function (x, y)
## c(x = 24 - 6 * y^2, y = -(y * (12 * x + 24 * y)))
\[ 24 - 6y^2 = 0 \\ 6y^2 = 24 \\ y^2 = 4 \\ y = \pm2 \]
\[ -12xy - 24y^2 = 0 \\ -12xy = 24y^2 \\ x = -2y \]
For y = 2, x = -4 For y = -2, x = 4
Substite in the original function to find critical points:
## [1] -64
## [1] 64
## # A tibble: 2 x 3
## x y z
## <dbl> <dbl> <dbl>
## 1 -4 2 -64
## 2 4 -2 64
Saddle Points (Inflection Points)
There are two inflection points, one is positive at D(-4,2) and the other is negative at D(4,-2)
Question 3
\[ Rev = price * units sold \\ R_1(x) = x * (81 - 21x + 17y) = (81x -21x^2 + 17xy) \\ R_2(y) = y * (40 + 11x -23y) = (40y + 11xy = 23y^2) \] Total Revenue \[ R(x,y) = 81x - 21x^2 + 17xy + 40y + 11xy -23y^2 \\ R(x,y) = -21x^2 + 28xy - 23y^2 + 81x +40y \]
## [1] 116.62
Question 4
$$ C(x,y) = x^2 + y^2 + 7x + 25y + 700 \ x + y = 96 \ \ C(96 - y, y) = (96 - y)^2 + y^2 + 7(96 - y) + 25y + 700 \ C(96 - y, y) = y^2 - 14y + 2908
$$
Minimum Value
\[ \frac{2}{3}y - 14 = 0 \\ y = 21 \\ x = 96 - 21 \\ x = 75 \] The company needs to produce 75 units in LA and 21 in Denver to minimize cost
Question 5
## $Q
## [1] 5.341559e+17
##
## $error
## [1] 15214781905