A waste management company is designing a rectangular construction dumpster that will be twice as long as it is wide and must hold 17 yd cubed of debris. Find the dimensions of the dumpster that will minimize its surface area.

x  =  4,67  yd

w  =  2,34  yd

h  =  1,56 yd

Step-by-step explanation:

Let call   x ,  w   and  h ( long, wide and high of the dumpster

and according to problem statement

x  =  2*w      and     V(d)  = x*w*h   and as x = 2w   ⇒ w = x/2

V(d)  = x*(x/2)*h     ⇒   17  = x²*h /2       ⇒  h  =  34/x²

Area of the dumpster is equal to area of the base plus sides areas ( dumpster is without lid)

Area of the base

A₁  = x*x/2   ⇒  A₁  = x²/2

We have four sides, two of them with base x  and to with base x/2

A₂  =  2* x*h   +  2* (x/2)*h          as  h  =34/x²

A₂  = 2* x*34/x²  + 2*( x/2)*34/x²     ⇒ 68/x  +  34/x

And total area as function of x is        A₁  +  A₂

A(x)  =  x²/2   +  102/x

Taking derivatives on both sides of the equation we have:

A´(x)  =  x  -  102/x²

A´(x)  =  0          ⇒      x  -  102/x²  =  0      ⇒ x³   - 102  = 0

x = ∛102

x  =  4,67  yd

w  =  4,67/2       ⇒   w  =  2,34  yd

h  = 34/(4,67)²    ⇒  h  =  1,56 yd

the probability is 0