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)
lm<- lm(y~x)
summary(lm)##
## 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-05The equation of the regression line for the given points is
\[y=4.26x-14.80\]
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.
\[f(x,y)=24x-6x{ y }^{ 2 }-8{ y }^{ 3 }\]
\({ f }_{ x }=\frac { df }{ dx } =24-6{ y }^{ 2 }\) \({ f }_{ y }=\frac { df }{ dy } =-12xy-24{ y }^{ 2 }\)
For critical points, put \({ f }_{ y }=0, and { f }_{ y }=0\), so
\(24-6{ y }^{ 2 }=0\quad \rightarrow y=\pm 2\)
\(-12xy-24{ y }^{ 2 }=0\quad \rightarrow x=-2y\) \(y=2\quad \rightarrow x=-4\) \(y=-2\quad \rightarrow x=4\)
The critical points are (-4,2) and (4, -2).
Now, we check for minima and maxima by taking second derivative
\({ f }_{ xx }=\frac { { d }^{ 2 }f }{ { dx }^{ 2 } } =0\) \({ f }_{ yy }=\frac { { d }^{ 2 }f }{ { dy }^{ 2 } } =-12x-48y\) \({ f }_{ xy }=\frac { { d }^{ 2 }f }{ dxdy } =-12y\)
Now, using the second derivative test discriminant:
\[D={ f_{ xx }{ f }_{ yy }-{ { f }_{ xy } }^{ 2 } }\] From the https://mathworld.wolfram.com/SecondDerivativeTest.html
So,
\(D=(0)*(-12x-48y)-{ (-12y) }^{ 2 }={ -144y }^{ 2 }\)
At point (-4,2),\(D=-144*(2)^2=-576 < 0\) and also at point (4, -2), \(D=-144*(-2)^2=-576 < 0\).
Therefore, both points (-4,2) and (4, -2) are saddle points. There are no local maxima and local minima in this case.
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).
Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?
Step 1:
Revenue = Units Sold x Sales Price
\(R(x,y) = x(81 − 21x + 17y) + y(40 + 11x - 23y)\)
\(R(x,y) = 81x − 21x^2 + 17xy + 40y + 11xy − 23y^2\)
\(R(x,y) = 81x + 40y + 28xy − 21x^2 − 23y^2\)
Step 2:
\(R(2.3,4.1) = 81*2.3 + 40*4.1 + 28*2.3*4.1 − 21*(2.3)^2 − 23*(4.1)^2\) \(R(2.3,4.1) = \$116.62\)
So, the revenue is $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?
\(C(x,y)=\frac { 1 }{ 6 } { x }^{ 2 }+\frac { 1 }{ 6 } { y }^{ 2 }+7x+25y+700\)
\(x + y = 96\)
Substitute \(y = 96 - x\) into the given total weekly cost equation.
\(C(x)=\frac { 1 }{ 6 } { x }^{ 2 }+\frac { 1 }{ 6 } { (96 - x) }^{ 2 }+7x+25(96 - x)+700\)
\(C(x)=\frac { 1 }{ 6 } { x }^{ 2 }+\frac { 1 }{ 6 } { (9216-192+{ x }^{ 2 }) }+7x+2400-25x+700\)
\(C(x)=\frac { 1 }{ 3 } { x }^{ 2 }-50x+4636\)
To minimize the total weekly cost, we take the derivative of C’(x)=0.
\(\frac { 2 }{ 3 } x-50=0\)
\(x=75\)
\(y=96-75=21\)
Therefore, the plant in Los Angeles should produce 75 units and the plant in Denver should produce 21 units to minimize the total weekly cost.
Evaluate the double integral on the given region.
\[\iint _{ R }^{ }{ ({ e }^{ 8x+3y })dA;R:2\le x\le 4\quad and\quad 2\le y\le 4 }\]
Write your answer in exact form without decimals.
\(\int _{ y=2 }^{ y=4 }{ \int _{ x=2 }^{ x=4 }{ ({ e }^{ 8x+3y })dxdy } }\)
\(=\int _{ y=2 }^{ y=4 }{ \int _{ x=2 }^{ x=4 }{ { (e }^{ 8x }{ e }^{ 3y })dxdy } }\)
\(=\int _{ 2 }^{ 4 }{ \frac { 1 }{ 8 } } { e }^{ 3y+32 }-\frac { 1 }{ 8 } { e }^{ 3y+16 }dy\)
\(=\int _{ 2 }^{ 4 }{ \frac { 1 }{ 8 } } ({ e }^{ 16 }-1){ e }^{ 3y+16 }dy\)
\(=\frac { e^{ 44 }-{ e }^{ 28 } }{ 24 } -\frac { e^{ 38 }-{ e }^{ 22 } }{ 24 }\)
\(=\frac { 1 }{ 24 } ({ e }^{ 22 }-{ e }^{ 28 }-{ e }^{ 38 }+{ e }^{ 44 })\)