The dimensions that will minimize the cost is
h= 2.82876 and r = 1.06078
Given That, C₁ = $10 and C₂= $10 and C₃ = $15
lets multiply the top and bottom, we get,
C₁ × C₂ = $10 ×$ 20 = $ 30
The total cost is,
C = (2πr²) C₁ + (2πrh)C₂
Here, r is base radius and h is the side length.
The volume is given by and the volume to 20 m³
∴ V = πrh² ⇒ πr²h = 20
So, h = 20/πr²
Now, the problem can be stated as
minimum h, r C(h, r) restricted to V(h, r) = V₀
Using language multiplier it reads,
L(h ,r,λ) = C(h, r) + λ [ V(h ,r) - V₀]
Solving for h, r, λ, we get,
h = ∛(2 C₁/C₂)² V₀ /π , r = ∛ (C₂/C₁) (v₀/2π)
λ = -2 ∛2πC₁C₂² /V₀
or, h= 2.82876 , r= 1.06078 with associated cost C= 424.214
We know that is a minimum because,
(C₀ V) = 2 ( C₁πr³ + C₂V₀) /r
d²/dr² (C₀V) (r) = 4 (C₁πr³+ C₂V₀)/ r³
and for the found solution has the value
d²/dr² (C₀V) (r) = 240 π > 0
Learn more about multiplier:
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