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
)
To find the equation of the line, I will just be using the summary() command after setting up all the points into a linear regression model.
x = c(5.6,6.3,7,7.7,8.4)
y = c(8.8,12.4,14.8,18.2,20.8)
lm_reg = lm(y~x)
summary(lm_reg)##
## 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-05For the summary, we can see the equation of the regression line for
the given points is:
\(y = -14.8x + 4.26\)
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.
\(f(x,y) = 24x - 6xy^2 - 8y^3\)
Find the 1st partial derivatives for x and y:
\(\frac{df}{dx} = 24 - 6y^2\)
\(\frac{df}{dy} = -(6*2)xy - (8*3)y^2 = -12xy
- 24y^2\)
Then we solve x and y for 0 to find our points:
\(0 = 24 - 6y^2\)
\(6y^2 = 24\)
\(y^2 = 4\)
\(y = +/- 2\)
Need to solve for x using both versions of y values:
\(0 = -12xy - 24y^2\)
\(y = -2\)
\(0 = -(12*-2)x - 24(-2)^2\)
\(0 = -(-24)x - 24(4)\)
\(0 = 24x - 96\)
\(-24x = -96\)
\(x = 4\)
\(y = 2\)
\(0 = -(12*2)x - 24(2)^2\)
\(0 = -(24)x - 24(4)\)
\(24x = -96\)
\(x = -4\)
So our two points are: \((4,-2),(-4,2)\)
Now we find 2nd partial derivatives:
\(\frac{d^2f}{dx} = 0\)
\(\frac{d^2f}{dy} = -12x - 48y\)
\(\frac{d^2f}{dxdy} = -12y\)
We need evaluate each derivative with both points:
(\(\frac{d^2f}{dx} = 0\) has no x or y
to plug into so we will ignore this one)
At \((4,-2)\):
\(\frac{d^2f}{dy} = -12x - 48y = -12(4) -
48(-2) = -48 + 96 = 48\)
\(\frac{d^2f}{dxdy} = -12y = -12(-2) =
24\)
At \((-4,2)\):
\(\frac{d^2f}{dy} = -12x - 48y = -12(-4) -
48(2) = 48 - 96 = -48\)
\(\frac{d^2f}{dxdy} = -12y = -12(2) =
-24\)
Now we plug into the 2nd derivative test to see if any of these are a
local maxima or minima:
The formula is apparently: \(D(x,y) =
(\frac{d^2f}{dx} * \frac{d^2f}{dy}) - \frac{d^2f}{dxdy}^2\)
\(D(4,-2) = (0 * 48) - 24^2 = 0 - 576 =
-576\)
\(D(-4,2) = (0 * -48) - (-24)^2 = 0 - 576 =
-576\)
Because they are both negative values, these are saddle points and we
have no local maxima or minima.
Our saddle points are: \((4,-2,-576),(-4,2,-576)\)
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).
Labeling the “house” brand as x and “name” brand as y:
\(R = x(81 - 21x + 17y) + y(40 + 11x -
23y)\)
Step 2. What is the revenue if she sells the “house” brand for $2.30
and the “name” brand for $4.10?
Solving for x = 2.3 and y = 4.1:
\(R = x(81 - 21x + 17y) + y(40 + 11x -
23y)\)
\(R = 2.3(81 - 21(2.3) + 17(4.1)) + 4.1(40 +
11(2.3) - 23(4.1))\)
\(R = 2.3(81 - 48.3 + 69.7) + 4.1(40 + 25.3 -
94.3)\)
\(R = 2.3(102.4) + 4.1(-29)\)
\(R = 235.52 - 118.9\)
\(R = 116.62\)
The revenue if she sells the “house” brand for $2.30 and the “name”
brand for $4.10 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?
\(x + y = 96\) units
We need to substitute either x or y for the other variable in the
equation. I am doing sub for x as I would rather work with the 7x than
the 25y so x = 96 - y:
\(C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 +
7x + 25y + 700\)
\(C(y) = \frac{1}{6}(96 - y)^2 +
\frac{1}{6}y^2 + 7(96 - y) + 25y + 700\)
\(C(y) = \frac{1}{6}(9216 - 192y + y^2) +
\frac{1}{6}y^2 + 672 - 7y + 25y + 700\)
\(C(y) = 1536 - 32y + \frac{1}{6}y^2 +
\frac{1}{6}y^2 + 672 - 7y + 25y + 700\)
\(C(y) = \frac{1}{3}y^2 - 14y +
2908\)
Take the C’(y) to get the min value for y:
\(C'(y) = \frac{2}{3}y - 14 =
0\)
\(C'(y) = \frac{2}{3}y = 14\)
\(C'(y) = y = 21\)
So for the min units for y/from Denver is 21.
We can then plug into our simple equation above to find min units for
x/from Los Angeles:
\(x + y = 96\)
\(x + 21 = 96\)
\(x = 75\)
To minimize the total weekly cost, Los Angeles needs to produce 75 units
while Denver produced 21 units.
Evaluate the double integral on the given region.
\(∫∫e^{8x + 3y} dA ; R: 2 ≤ x ≤ 4 and 2 ≤ y ≤
4\)
Write your answer in exact form without decimals.
\(2∫^42∫^4e^{8x + 3y} dy dx\) (the 2s
are the denom for ∫)
\(2∫^4 (\frac{1}{3}e^{8x + 3y}| 4, 2)
dx\)
\(2∫^4 (\frac{1}{3}e^{8x + (3*4)} -
\frac{1}{3}e^{8x + (3*2)}) dx\)
\(2∫^4 (\frac{1}{3}e^{8x + 6} (e^6-1))
dx\)
\((\frac{1}{24}e^{8x + 6} (e^6-1)) | 4,
2\) \((\frac{1}{24}e^{8*4 + 6} (e^6-1))
- (\frac{1}{24}e^{8*2 + 6} (e^6-1))\)
\((\frac{1}{24}e^{38} (e^6-1)) -
(\frac{1}{24}e^{22} (e^6-1))\)
\(\frac{1}{24}e^{38}-e^{22}
(e^6-1)\)
\(\frac{1}{24}(e^{44}-e^{28}-e^{38}+e^{22})\)