Answer:
Dimensions of the container:
x = 3 m
y = 6 m
h = 1.1 m
C(min) = 270 $
Step-by-step explanation:
Volume of storage container
V = 20 m³
Let "y" be the length and "x" the width then y = 2*x
V = x*y*h ⇒ V = 2*x²*h ⇒ 20 = 2*x²+h ⇒ h = 10/ x²
Costs:
Total cost = cost of base ( 5*2*x² ) + cost of side with base x ( 2*9*x*h) +
cost of side witn base y =2x (2*9*2x*h)
C(t) = 10*x² + 18*x*h + 36*x*h
C(x) = 10x² + 54*x*10/x² ⇒ C(x) 10*x² + 540 /x
Taking derivatives on both sides of the equation we get:
C´(x) = 20*x - 540/x²
C´(x) = 0 ⇒ 20*x - 540/x² = 0 ⇒ 2x - 54/x² = 0
2x³ - 54 = 0
x³ = 27 x = 3 m
Then y = 2*x ⇒ y = 2*3 y = 6 and h = 10 / x² h = 1.1 m
And the minimum cost is
C (min) = 10*x² + 540/x ⇒ C (min) = 90 + 180
C(min) = 270 $
So you can do this multiple ways, I'll do this the way that I think makes sense the l most easily.
Cos (0) = 1
Cos (pi/2)=0
Cos (pi) =-1
Cos (3pi/2)=0
Cos (2pi)=1
Now if you multiply the inside by 4, the graph oscillates more violently (goes up and down more in a shorter period).
But you can always reduce it.
Cos (0)= 1
Cos (4pi/2) = cos (2pi)=1
Cos (4pi) =Cos (2pi) =1 (Any multiple of 2pi ==1)
etc...
the pattern is that every half pi increase is now a full period as apposed to just a quarter of one. That's in theory.
Now that you know that, the identities of Cosine are another beast, but mathematically.
You have.
Cos (2×2t) = Cos^2 (2t)-Sin^2 (2t)
Sin^2 (t)=-Cos^2 (t)+1..... (all A^2+B^2=C^2)
Cos (2×2t) = Cos^2 (t)-(-Cos^2 (t)+1)
Cos (2×2t)= 2Cos^2 (2t) - 1
2Cos^2 (2t) -1= 2 (Cos^2(t)-Sin^2(t))^2 -1
(same thing as above but done twice because it's cos ^2 now)
convert sin^2
2Cos^2 (2t)-1 =2 (Cos^2 (t)+Cos^2 (t)-1)^2 -1
2 (2Cos^2(t)-1)^2 -1
2 (2Cos^2 (t)-1)(2Cos^2 (t)-1)-1
2 (4Cos^4 (t) - 2 (2Cos^2 (t))+1)-1
Distribute
8Cos^4 (t) -8Cos^2 (t) +1
Cos (4t) =8Cos^4-8Cos^2 (t)+-1
Answer:
First do 6+7 (PEMDAS), plus 2 and then plus 54
2+(-11) is -9
Sec(theta) = 1 / cos(theta) = 5 / 3 solve for cos(theta)
The actual factor is 6(x + 1)(x^2 - x +1).