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

f(x) is going down by 4

step-by-step explanation:

(its either that or its going left by 4)