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 )
\[ \frac {\sum_{i=1}^n (x_i -\bar{x}) (y_i -\bar {y})} {\sum_{i=1}^n (x_i -\bar{x}) ^2}\\ \]
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
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 dollars?
House Brand \[ 81-21x+17y \] Name Brand \[ 40-11x-23y \] Revenue Function R(x,y) \[ R(x,y) = x(81-21x+17y) + y(40+11x-23y)\\ =-21x^2 - 23y^2+28xy+81x+40y \] Step 2, The amount of Revenue if house brand is $2.30 per unit & name brand is 4.10 dollars. \[ R(x,y) =-21x^2 - 23y^2+28xy+81x+40y\\ R(2.3,4.1) \Rightarrow -21(2.3)^2 -23(4.1)^2 +28(2.3)(4.1)+81(2.3)+40(4.1)\\ \Rightarrow 116.62 \ dollars \]
a= -21*(2.3)^2 -23*(4.1)^2 +28*(2.3)*(4.1)+81*(2.3)+40*(4.1)
a## [1] 116.62A 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) = 1/6 x2 + 1 6 y2 + 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?
Evaluate the double integral on the given region. \[ \iint_R (e^{8x+3y})dA; R:2 \leq x \leq 4\ and\ 2 \leq x \leq 4\ \\ = \int_2^4 \bigg[ \int_2^4 e^{8x+3y}dx \bigg] dy\\ =\int_2^4 \bigg [\int_2^4 e^{8x} e^{3y}dx \bigg]dy \\ = \int_2^4 e^{3y} \bigg[ \int_2^4 e^{8x}dx \bigg] dy\\ =\int_2^4 e^{3y} \bigg[ \frac{1}{8} e^{8x} \bigg|_2^4 \ \bigg] dy\\ = \int_2^4 e^{3y} \bigg[ \frac {e^{32} - e^{16}} {8} \ \bigg] dy\\ =\bigg[ \frac {e^{32} - e^{16}} {8} \ \bigg] \int_2^4 e^{3y}dy\\ =\bigg( \frac {e^{32} - e^{16}} {8} \ \bigg) \bigg[\frac{1}{3} e^{3y} \bigg|_2^4\bigg] \\ =\bigg( \frac {e^{32} - e^{16}} {8} \ \bigg) \bigg(\frac {e^{12} -e^6}{3} \bigg) \]