The base and sides of a container is made of wood panels. The container does not have a lid. The base and sides are rectangular. The width of the container is x c m. The length is fourth times the width. The volume of the container is 400 c m 3. Determine the minimum surface area that this container will have.

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