Answer:
500
Step-by-step explanation:
Answer is 2/ x2/3. I hope this worked. Good luck.
For the answer to the question above,
<span>r = 1 + cos θ
x = r cos θ
x = ( 1 + cos θ) cos θ
x = cos θ + cos^2 θ
dx/dθ = -sin θ + 2 cos θ (-sin θ)
dx/dθ = -sin θ - 2 cos θ sin θ
y = r sin θ
y = (1 + cos θ) sin θ
y = sin θ + cos θ sin θ
dy/dθ = cos θ - sin^2 θ + cos^2 θ
dy/dx = (dy/dθ) / (dx/dθ)
dy/dx = (cos θ - sin^2 θ + cos^2 θ)/ (-sin θ - 2 cos θ sin θ)
For horizontal tangent line, dy/dθ = 0
cos θ - sin^2 θ + cos^2 θ = 0
cos θ - (1-cos^2 θ) + cos^2 θ = 0
cos θ -1 + 2 cos^2 θ = 0
2 cos^2 θ + cos θ -1 = 0
Let y = cos θ
2y^2+y-1=0
2y^2+2y-y-1=0
2y(y+1)-1(y+1)=0
(y+1)(2y-1)=0
y=-1
y=1/2
cos θ =-1
θ = π
cos θ =1/2
θ = π/3 , 5π/3
θ = π/3 , π, 5π/3
when θ = π/3, r = 3/2
when θ = π, r = 0
when θ = 5π/3 , r = 3/2
(3/2, π/3) and (3/2, 5π/3) give horizontal tangent lines
</span>---------------------------------------------------------------------------------
For horizontal tangent line, dx/dθ = 0
<span>-sin θ - 2 cos θ sin θ = 0 </span>
<span>-sin θ (1+ 2 cos θ ) = 0 </span>
<span>sin θ = 0 </span>
<span>θ = 0, π </span>
<span>(1+ 2 cos θ ) =0 </span>
<span>cos θ =-1/2 </span>
<span>θ = 2π/3 </span>
<span>θ = 4π/3 </span>
<span>θ = 0, 2π/3 ,π, 4π/3 </span>
<span>when θ = 0, r=2 </span>
<span>when θ = 2π/3, r=1/2 </span>
<span>when θ = π, r=0 </span>
<span>when θ = 4π/3 , r=1/2 </span>
<span>(2,0) , (1/2, 2π/3) , (0, π), (1/2, 4π/3) </span>
<span>At (2,0) there is a vertical tangent line</span>
S+m+l=28
4s+2m+l=58
6s+5m+4l=135
Eliminate the variable l from the first 2 equations
s+m+l=28
-4s-2m-l=-58
-3s-m=-30
Elminiate the variable l from the last 2 equations
6s+5m+4l=135
-16s-8m-4l=-232
-10s-3m=-97
Now solve for s and m using the 2 equations without l
-3s-m=-30
-10s-3m=-97
9s+3m=90
-10s-3m=-97
-s=-7
s=7
Then plug in s into one of the equations without l
-3(7)-m=-30
-21-m=-30
-m=-9
m=9
Now plus in s and m into one of the original 3 equations
(7)+(9)+l=28
16+l=28
l=12
Final answer:
Small=$7
Medium=$9
Large=$12
I know it only asks for large but I wanted to show you how to find them all for future reference. :)