Home Work 15

#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)

b0 <- round(summary(lm(y ~ x))$coefficients[1], 2)
b0
## [1] -14.8
b1 <- round(summary(lm(y ~ x))$coefficients[2], 2)
b1
## [1] 4.26
paste0("The equation of the regression line is y = ", b0 , " + ", b1 , " x")
## [1] "The equation of the regression line is y = -14.8 + 4.26 x"

#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.

\[f ( x, y ) = 24x - 6xy^2 - 8y^3 \]

\[\frac{df}{dx}=24−6y^2\]

\[\frac{df}{dy}=−12xy−24y^2\]

Set \(fx(x, y) = 0\) to find value of y
\[ 0=24−6y^2 \ ; \ y^2=4 \ ; \ y=±2\]

Set \(fy(x, y) = 0\) and use valye of y from above to find value of x
\[ y=±2 \ and \ −12xy−24y^2=0 \ ; \ ±24x=24x4 \ ; \ x=±4 \]

#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 \(R ( x, y )\).

\[Revenue = (Units Sold) x (Sales Price) \] \[R(x,y)=(x∗(81−21x+17y))+(y∗(40+11x−23y)) \] \[R(x,y)=81x−21x^2+17xy+40y+11xy−23y^2 \] \[R(x,y)=−21x^2−23y^2+81x+40y+28xy \]

Step 2. What is the revenue if she sells the “house” brand for $2.30 and the “name” brand for $4.10?

x=2.30
y=4.10

revenue <- -21*x^2 +81*x +28*x*y +40*y -23*y^2

paste0("The revenue is = ", round(revenue, 4))
## [1] "The revenue is = 116.62"

#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?

==> we know \(x+y=96\) and therefore \(y=96−x\) ; so if change the cost function to \(C(x,y)=C(x,96−x)\), then we get

\[=\frac{1}{6}x^2+\frac{1}{6}(96−x)^2+7x+25(96−x)+700\]

this then simplifies to \[=\frac{1}{3}x2−50x+4636\]

Derivative of the Cost Function and Local Minimum we get \[C′=(\frac{2}{3})x−50=0 \] \[x = 75 \]

since \(x+y=96\) ; then \(y=21\)

Hence 75 units produced in Los Angeles and 21 in Denver is optimal to have the minimum cost.

#5. Evaluate the double integral on the given region.

\[ \int\int_R e^{8x + 3y} dA ; R: 2 \le x \le 4 \ and \ 2 \le y \le 4 \]

Write your answer in exact form without decimals.

\[\int_2^4 \int_2^4e^{8x+3y}dxdy \]

\[\int_2^4(\int_2^4(e^{8x+3y})dx)dy \]

\[= \int_2^4e^{3y}(\int_2^4(e^{8x})dx)dy \]

\[= \int_2^4e^{3y}((1/8)e^{8x})|_2^4dy \]

\[= \int_2^4(e^{3y}/8)(e^{32}−e^{16})dy \]

\[= ((e^{32}−e^{16})/8)\int_2^4e^{3y}dy \]

\[= ((e^{32}−e^{16})/8)((1/3)e^{3y})|_2^4 \]

\[= (1/24)(e^{32}−e^{16})(e^{12}−e^6) \]

eval <- 1/24*(exp(32)-exp(16))*(exp(12)-exp(6))
eval
## [1] 5.341559e+17
paste0("Exact form without decimals is : ", format(eval, scientific = F))
## [1] "Exact form without decimals is : 534155947497085120"