Answer:
A(mim ) = 300 cm²
Step-by-step explanation:
Area of the base is:
A = 4*x*x ⇒ A = 4*x²
Let call h the height of the container, then area lateral are:
A₁ = 2*4*x*h A₁ = 8*x*h and
A₂ = 2* x * h A₂ = 2*x*h
From the volume of the container we have:
400 = Area of the base*h
400 = 4*x²*h ⇒ h = 100 /x²
Now Total area of the container is :
A = 4*x² + 8*x*h + 2*x*h ⇒ A = 4*x² +10*x*h
As h = 100/x²
A(x) = 4*x² + 10*x* 100/x²
A(x) = 4*x² + 1000/x
Taking derivatives on both sides of the equation we get:
A´(x) = 8*x - 1000/x²
A´(x) = 0 ⇒ 8*x - 1000/x² = 0
8*x³ = 1000 = 0 ⇒ x³ = 1000 / 8 ⇒ x = ∛(1000)/8
x = 5 cm
Now minimum area is:
Area of the base
4*5*5 = 100 cm²
A₁ = 8*x*h h = 100 / x² h = 100 / 25 h = 4 cm
A₁ = 8*5*4
A₁ = 160 cm²
A₂ = 2*5*4
A₂ = 40 cm²
A(mim ) = 300 cm²
