data_605_hw15
Question 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)
df = data.frame(x, y)
model <- lm(y ~ x, data = df)
summary(model)##
## Call:
## lm(formula = y ~ x, data = df)
##
## 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-05model$coefficients## (Intercept) x
## -14.800000 4.257143\[y = -14.80 + 4.28x\]
Question 2
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.
\[f(x,y)=24x-6xy^2-8y^3\]
First, we need to compute the partial derivatives with respect to x and y. We begin by ordering anything that is not an x, to the front for each term/factor.
\[ f(x,y)=24x-y^{2}6x-8y^{3}\\ f_{x}(x, y)=24-6y^{2} \] \[f_y(x,y) = -12xy - 24y^2\]
Now, we can set each partial derivative equal to zero and solve the resulting system of two equations and two unknowns.
If \(24 - 6y^2 = 0\), then \(y^2 = 4\) and thus \(y \pm 2\).
If \(y = 2\) and \(-12xy - 24y^2 = 0\), then \(-24x = 24(2^2)\) and thus \(x = -4\).
If \(y = -2\) and \(-12xy - 24y^2 = 0\), then \(24x = 24(2^2)\) and thus \(x = 4\).
Now, calculate \(f(x,y)\).
\[f(4,-2) = 24(4) - 6(4)(-2^2) - 8(-2^3)\] \[= 96 - 96 + 64\] \[f(4,-2) = 64\] \[f(-4,2) = 24(-4) - 6(-4)(2^2) - 8(2^3)\] \[= -96 + 96 - 64\] \[f(-4,2) = -64\]
Two Critical Points \((4, -2, 64)\) and \((-4, 2, -64)\)
Now, we can use the Second Derivative Test to determine if the points are at local maxima, local minima, or saddle points. A saddle point is where a critical point exists but does not have any extrema.
Second Derivatives
\(f_{xx} = 0\) and \(f_{yy} = -12x - 48y\) and \(f_{xy} = -12y\)
Consider
\[D(x,y) = f_{xx}f_{yy} - f_{xy}^2\] \[= (0)(-12x - 48y) - (-12y)^2\] \[= -144y^2\] \(D(x,y) < 0\) for all \((x, y)\), thus any critical point is a saddle points.
Therefore both critical points \((4, -2)\) and \((-4, 2)\) are saddle points.
Question 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.
Step 1. Find the revenue function R(x,y). \[ R(x,y) = (81- 21x + 17y)x + (40 + 11x - 23y)y\\ R(x, y)=81x - 21x^2 + 28yx + 40y - 23y^2 \]
Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?
house <- 2.3
name <- 4.1
revenue <- (81 * house) - (21 * house^2) + (28 * name * house) + (40 * name) - (23 * name^2)
print(glue("The revenue is roughly $", {revenue}))## The revenue is roughly $116.62Question 4
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?
\[ x + y = 96\\ x = 96 - y \]
We can convert C(x, y) into a univariate function. \[ c(x, y)=\frac{1}{6}x^{2}+\frac{1}{6}y^{2}+7x+25y+700\\ =\frac{1}{6}(96-y)^{2}+\frac{1}{6}y^{2}+7(96-y)+25y+700\\ C(y)=\frac{1}{3}y^{2}-14y+2908 \]
Let’s take the first derivative to find the minimal value. \[ C'(y)=\frac{2}{3}y-14=0\\ \frac{2}{3}y=14\\ 2y=42\\ y=21 \]
Substitute it to find x value. \[ x = 96-y\\ x = 96-21\\ x = 75 \] We need 75 units from Los Angeles and 21 units from Denver in order to minimize the cost.
Question 5
Evaluate the double integral on the given region.
\[\iint (e^{8x+3y})\ dA;R:2 \leq x \leq 4\ and\ 2 \leq y \leq 4\]
Write your answer in exact form without decimals.
\[\int_2^4 \int_2^4 (e^{8x+3y})\ dy\ dx\] \[= \int_2^4 \bigg(\bigg[\frac{1}{3}e^{8x+3y}\bigg]\bigg|_2^4\bigg)\ dx\] \[= \int_2^4 \bigg(\bigg[\frac{1}{3}e^{8x+12}\bigg]-\bigg[\frac{1}{3}e^{8x+6}\bigg]\bigg)\ dx\] \[= \int_2^4 \bigg(\frac{1}{3}e^{8x+6}\bigg[e^{6}-1\bigg]\bigg)\ dx\] \[= \bigg(\frac{1}{24}e^{8x+6}\bigg[e^{6}-1\bigg]\bigg)\bigg|_2^4\]
\[= \frac{1}{24}e^{38}(e^{6}-1)-\frac{1}{24}e^{22}(e^{6}-1)\] \[= \frac{1}{24}(e^{38}-e^{22)}(e^{6}-1)\] \[= \frac{1}{24}(e^{44}-e^{38}-e^{28}+e^{22})\]