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 $
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<em>Honestly, I've been in </em><em>a</em><em> </em><em>similar</em><em> situation before, that's how I know what to do.</em>
Answer:
length TR = 6
Step-by-step explanation:
Length TL = 24
scale ratio = 3:4:5
T-------R--------V----------L
find: TR
24 / (3+4+5) = 2
length TR at 3 x 2 = 6
length RV at 4 x 2 = 8
length VL at 5 x 2 = 10
a total of 6 + 8 + 10 = 24
therefore,
length TR = 6
