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 )
By some proofs I’ve long forgoten, the least squares solution can be determined by \(\hat { x } ={ ({ A }^{ T }A) }^{ -1 }{ A }^{ T }b\).
A <- c(rep(1, 5), 5.6, 6.3, 7, 7.7, 8.4)
b <- c(8.8, 12.4, 14.8, 18.2, 20.8)
dim(A) <- c(5,2)
solve(t(A) %*% A) %*% t(A) %*% b## [,1]
## [1,] -14.800000
## [2,] 4.257143x <- (A[, 2])
df <- data.frame(X = x, y = b)
lm(y ~ x, data = df)##
## Call:
## lm(formula = y ~ x, data = df)
##
## Coefficients:
## (Intercept) x
## -14.800 4.257We see this matches the solution given by the lm function
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.
\(f(x,y)=24x-6x{ y }^{ 2 }-8{ y }^{ 3 }\)
\({ f }_{ x }=24-6{ y }^{ 2 }, { f }_{ xx }=0\)
\({ f }_{ y }=12xy-24{ y }^{ 2 },{ f }_{ yy }=12x-48{ y }\)
\({ f }_{ xy }=-12{ y }\)
\(24-6{ y }^{ 2 }=0,\quad 12xy-24{ y }^{ 2 }=0\)
\(y=-2,2\)
at \(y = 2\), so \(24x=96\)
at \(y = -2\), so \(-24x=96\)
\(x = -4,4\)
By the second derivative test since \({ f }_{ xx }\) is zero, the test will be positive, so (-4,-2,64) and (4,2,-64) are both saddle points
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*
We know \({ Q }_{ s }=81-21x+17y,{ Q }_{ n }=40+11x-23y\) so, \({ R }(x,y)=x(81-21x+17y)+y(40+11x-23y)\)
Step 2
\({ R }(2.3,4.1)=2.3(81-21*2.3+17*4.1)+4.1(40+11*2.3-23*4.1)\)
=$91.04 However since quantity can’t be negative the function is really \({ R }(2.3,4.1)=max(2.3(81-2*2.3+17),0)+max(4.1(40+11*2.3-23*4.1),0)\)
with yields $233.52
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…
\(C(x,y)=1/6{ x }^{ 2 }+1/6{ y }^{ 2 }+7x+25y+700\) contrained by \(x+y=96\)
where \(x>=0\) and \(y>=0\)
\({ f }_{ x }=32x+7,\quad { f }_{ y }=32y+25\) So the one critical point occurs at \((-\frac { 7 }{ 32 } ,-\frac { 25 }{ 32 } )\) \(C(x,96-x)=16{ x }^{ 2 }+16(96-x)^{ 2 }+7x+25(96-x)+700\)
ignoring constants: \((x,96-x)=1/3{ x }^{ 2 }-32x+7x-25x\)
\({ (x,96-x) }^{ \prime }=\quad 2/3{ x }-50\)
\(x=75,y=21\) represents a critical point
\(f(75,21)=2761\)
\(f(96,0)=2908\)
\(f(0,96)=4636\)
Cost is minimized on that interval at (75,21)
\(\int _{ 2 }^{ 4 } \int _{ 2 }^{ 4 }{ { e }^{ 8x+3y } } dxdy\)
\(\frac { 1 }{ 8 } \int _{ 2 }^{ 4 }{ { e }^{ 8x+3y } } \overset { 4 }{ \underset { 2 }{ | } } dy\)
\(\frac { 1 }{ 8 } \int _{ 2 }^{ 4 }{ { e }^{ 8x }{ e }^{ 3y } } \overset { 4 }{ \underset { 2 }{ | } } dy\)
\(\frac { 1 }{ 8 } \int _{ 2 }^{ 4 }{ { e }^{ 3y } } ({ e }^{ 32 }-{ e }^{ 16 })dy\)
\(\frac { ({ e }^{ 32 }-{ e }^{ 16 }) }{ 24 } { { e }^{ 3y } }\overset { 4 }{ \underset { 2 }{ | } }\)
\(\frac { ({ e }^{ 32 }-{ e }^{ 16 }) }{ 24 } ({ e }^{ 12 }-{ e }^{ 6 })\)
5.341559510^{17}