The purpose of the assignment was to explore Calculus: Functions of Several Variables.
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)
Per R calculations below, the equation of the regression line is: \(y = -14.8 + 4.26x\)
##
## 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
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.
\(f(x,y) = 24x -6xy^2-8y^3\)
We start by taking the first derivative with respect to x and y to solve for critical points:
Setting \(f_x\) equal to 0 and solving for y, we get:
Setting \(f_y\) equal to 0 and substituting y = 2, we get:
Setting \(f_y\) equal to 0 and substituting y = -2, we get:
Thus, our critical points are (-4, 2) and (4, -2).
Next, we perform the second derivative test to determine whether we’re dealing with a saddle point (H < 0) or maximum / minimum (H > 0).
We derive \(f_{xx}\), \(f_{yy}\), \(f_{xy}\):
\(f_x = 24 - 6y^2\)
\(f_{xx} = 0\)
\(f_y = -12xy -24y^2\)
\(f_{yy} = -12x - 48y\)
\(f_{xy} = -12y\)
We substitute our expressions in for our H equation:
And then we substitute our critical point values (-4, 2) and (4, -2) to evaluate our H value:
Being that our y value is squared in the H expression above, it doesn’t matter whether our y value is positive or negative. The resulting H value is negative and thus we’ve identified both critical points (-4, 2) and (4, -2) as saddle points.
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.
$
2.30 and the “name” brand for $
4.10?The revenue function is simply the cost of the unit multiplied by the number of units sold for the “house” vs. “name” brand. It can be defined as:
\(R(x,y) = x(81 - 21x + 17y) + y(40 + 11x -23y)\)
When we substitute the provided values for x and y, we arrive at a revenue mark of $116.62 (as shown below):
## [1] 116.62
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 \(C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 +7x + 25y + 700\), 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?
We start by defining what we know:
We redefine our C(x,y) function as a single variable function C(x) by substituting for y:
At this point we want to find our critical points, so we take the derivative with regard to x, set it equal to 0 and solve for x and y:
To minimize weekly cost, x = 75 and y = 21. 75 units should be produced in LA and 21 should be produced in Denver.
To verify that it is indeed the minimum, we take the second derivative and verify that it’s positive.
The second derivative is positive. Thus, we’re dealing with a minimum point :)
Reference: https://www.mathsisfun.com/calculus/maxima-minima.html
Evaluate the double integral on the given region.
∬\(e^{8x+3y}dA\) for 2 <= x, y <= 4.
Write your answer in exact form without decimals.
We use the integral2() function from the “pracma” library to evaluate our double integrate and then compare to a by hand evaluation where we just plug in our values (2<= x, y <= 4). Our evaluation produces the following answer:
library(pracma)
fct <- function(x,y){
exp(8*x + 3*y)
}
i <- integral2(fct, 2, 4, 2, 4)
#display without decimals in our answer
print(i$Q, digits=17)
## [1] 534155947871807104
#Evaluating our double integral by hand and plugging in values gives:
check <- (1/3 * (exp(12) - exp(6))) * (1/8 * (exp(32)-exp(16)))
print(check, digits=17)
## [1] 534155947497085120
Being that the use of the builtin function is likely more accurate (machine > human for computation), we’ll use 534155947871807104.