Answer:
x = ∛ 2*V/5
y = ∛ 2*V/5
h = V/ ∛ 4*V²/25
Step-by-step explanation:
Dimensions of the aquarium base is x*y
We call c₁ cost per unit area of the sides, then cost per unit area of slate is equal 5c₁.
let call h the height of the aquarium then volume of the aquarium is:
V = x*y*h where h = V / x*y
As the base is a rectangular one there are 2 sides x*h . and 2 sides y*h
According to this:
Ct (cost of aquarium ) = cost of the base + cost of the sides
cₐ ( cost of the base) = 5*c₁*x*y
c₆ (cost of the sides ) = c₁*2*x*h + c₁*2*y*h
C(t) = 5*c₁*x*y +2* c₁*x* V/x*y + 2* c₁*y* V/x*y or
C(t) = 5*c₁*x*y + 2*c₁*V/y *2*c₁* V/x
Taking partial derivatives en x and y we have:
C´(x) = 5*c₁*y - 2*c₁*V/x²
C´(y) = 5*c₁*x - 2*c₁*V/y²
C´(x) = C´(y) ⇒ 5*c₁*y - 2*c₁*V/x² = 5*c₁*x - 2*c₁*V/y²
or 5*y - 2*V/x² = 5*x - 2*V/y²
(5*y*x² - 2*V)/x² = ( 5*y²x - 2*V) /y²
(5*y*x² - 2*V)*y² = ( 5*y²x - 2*V)*x²
5*y³*x² - 2*V*y² = 5*y²x³ - 2*V*x²
5*y³*x² - 5*y²x³ = 2*V * ( y² - x²)
by symmetry x = y
Then using x = y and plugging that value on the derivatives
C´(x) = 5*c₁*y - 2*c₁*V/x²
C´(x) = 5*c₁*x - 2*c₁*V/x²
C´(x) = 0 ⇒ 5*c₁*x - 2*c₁*V/x² = 0
5*x - 2*V/x² = 0 ⇒ 5*x³ - 2*V = 0 ⇒ 5*x³ = 2*V ⇒ x³ = 2*V/5
x = ∛ 2*V/5 and y = ∛ 2*V/5 and h = V/ x*y h = V/ ∛ 4*V²/25
→ 3 ≠ 1 → A is eliminated.
→ 5 ≠
→ B is eliminated.
→ 2 = 2 → C satisfies the equation.
→ 10 ≠
→ D is eliminated.
→ -1 = -1 → E satisfies the equation.
→ 3 = 3 → F satisfies the equation.