df = data.frame(rbind(c( 5.6, 8.8 ), c( 6.3, 12.4 ), c( 7, 14.8 ), c( 7.7, 18.2 ), c( 8.4, 20.8 )))
model <- lm(df$X2 ~ df$X1, df)
plot(df$X2 ~ df$X1, data=df)
abline(model)
Write your answer(s) in the form ( x, y, z ). Separate multiple points with a comma.
Solution: in order to find the answer we need the first and the second partial derivatives.
df/dx=24−6y^2 df/dy=−12xy−24y^2
d2f/dx=0 d2f/dy=−12x−48y
df/dx=24−6y2=0−>4−y2=0 df/dy=−12xy−24y2=0−>−xy−2y2=0
y=+/-2. when y=2 then x=-4 and when y=-2 then x=4. So our points are (4,-2) and (-4,2).
for (4,-2) f(x,y)=24∗4−6∗4∗(−2)2−8(−2)3=64 which is > 0
for (-4,2) f(x,y)=24∗−4−(6∗−4∗(2)2)−8(2)3=−64 which is < 0
So only (-4,2) is the saddle point.
House brand: R(x)=x∗(81−21x+17y)
Name brand: R(y)=y∗(40+11x−23y)
Total = $R(x,y)=x∗(81−21x+17y)+y∗(40+11x−23y) -> - 21x^2 - 23y^2 + 28xy + 81x + 40y $
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
total <- -21*x^2 - 23*y^2 + 28*x*y + 81*x + 40*y
print(total)
## [1] 116.62
x+y=96 so we substitute y=96−x
reducing the C(x,y) gives us x2−50∗x+4636
Solving the first derivative gives us x=75 which means that y=21. LA must produce 75 units and Denver 21 to reduce the total weekly costs.
Write your answer in exact form without decimals.
e(8x+3y)=e(8∗x)+e(3∗y)
evaluating each section individually you get
(e12−e6)(e32−e16)/24