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 )

ref_site

sum_xy = 5.6 * 8.8 + 6.3 * 12.4 + 7 * 14.8 + 7.7 * 18.2 + 8.4 * 20.8
sum_x = 5.6 + 6.3 + 7 + 7.7 + 8.4
sum_x2 = 5.6^2 + 6.3^2 + 7^2 + 7.7^2 + 8.4^2
sum_y = 8.8 + 12.4 + 14.8 + 18.2 + 20.8
sum_y2 = 8.8^2 + 12.4^2 + 14.8^2 + 18.2^2 + 20.8^2

s2_x = (sum_x2 - sum_x2/5)/4
s2_y = (sum_y2 - sum_y2/5)/4

sd_x = sqrt(s2_x)
sd_y = sqrt(s2_y)

s_xy = (sum_xy - (sum_x - sum_y)/5)/4
r = s_xy/(s2_x*s2_y)
b = r*(s2_y/s2_x)
y_int = sum_y/5 - b*sum_x

b
## [1] 0.05543034
y_int
## [1] 13.05994

\(y = 0.06x + 13.06\)

  1. 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.

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

\(f(x) = 24 - 6y^2\)

$ y = $ \(f(y) = - 12xy - 24y^2\) \(f(y) = \pm 24x - 96\)

\(x = \pm4\)

\(z = 24(-4) - 6(-4)2^2 -8(2)^3 = -64\) \(z = 24(4) - 6(4)(-2)^2 -8(-2)^3 = 64\)

\((-4, 2, -64)\) \((4,-2, 64)\)

  1. 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 ).

\(R(x,y) = (81 - 21x + 17y)x + (40 + 11x - 23y)y\) \(R(x,y) = (81x - 21x^{2} + 17xy) + (40y + 11xy - 23y^{2})\) \(R(x,y) =81x - 21x^{2} + 28yx + 40y - 23y^{2}\)

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

x = 2.3
y = 4.1

81*x - 21*(x^2) + 28*y*x + 40*y - 23*(y^2)
## [1] 116.62
  1. 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?

\(x + y = 96\) \(y = 96 - x\)

#Substituting the value of y=96 - x into:

\(C(x, y) = \frac{1}{6}x^{2} + \frac{1}{6}y^{2} + 7x + 25y + 700\) \(C(x, y = 96 - x) = \frac{1}{6}x^{2} + \frac{1}{6}(96 - x)^{2} + 7x + 25(96 - x) + 700\) \(C(x, y = 96 - x) \frac{1}{6}x^2 + 1536 -32x +\frac{1}{6}x^2 + 7x + 2400 -25x + 700\) \(C(x, y = 96 - x) \frac{1}{3}x^2 -50x + 4636\)

  1. Evaluate the double integral on the given region.

I got some online help on this one.

\(\int\int (e^{8x + 3y}) dA ; R: 2 < x < 4 and 2 < y < 4\)

\(\int\int (e^{8x + 3y}) = \int_2^4 \left[ \frac{1}{8} \int_2^4 (e^{8x+3y}) \ 8 \ dx \right] \ dy\)

\(\int\int (e^{8x + 3y}) = \int_2^4 \left[ \left. \frac{1}{8} (e^{8x+3y}) \right|_2^4 \right] \ dy\)

\(\int\int (e^{8x + 3y}) = \int_2^4 \left[ \frac{1}{8} \left[ (e^{8(4)+3y}) - (e^{8(2)+3y})\right] \right] \ dy\)

\(\int\int (e^{8x + 3y}) = \int_2^4 \left[ \frac{1}{8} (e^{32}e^{3y}) - (e^{16} e^{3y}) \right] \ dy\)

\(\int\int (e^{8x + 3y}) = \int_2^4 \left[ \frac{1}{8} (e^{32} - e^{16}) e^{3y} \right] \ dy\)

\(\int\int (e^{8x + 3y}) = \frac{1}{8} (e^{32} - e^{16}) \int_2^4 \frac{1}{3} \left[ e^{3y} \right] 3 \ dy\)

\(\int\int (e^{8x + 3y}) = \frac{1}{8} (e^{32} - e^{16}) \frac{1}{3} \left[ \left. e^{3y} \right|_2^4 \right]\)

\(\int\int (e^{8x + 3y}) = \frac{1}{8} (e^{32} - e^{16}) \frac{1}{3} \left( e^{3(4)} - e^{3(2)} \right)\)

\(\int\int (e^{8x + 3y}) = \frac{1}{24} (e^{32} - e^{16}) \left( e^{12} - e^6 \right)\)

\(\int\int (e^{8x + 3y}) = \frac{1}{24} (e^{44} - e^{38} - e^{28} + e^{22})\)