A rectangular package sent by a postal service can have a maximum combined length and girth (perimeter of a cross sectio) of 108

Question

3 years ago

14

A rectangular package sent by a postal service can have a maximum combined length and girth (perimeter of a cross sectio) of 108

inches. Find the dimensions of the package of maximum volume that can be sent, if the cross sectional region must be square.

Mathematics

2 answers:

asambeis [7] · Answer 1 · 2022-04-19T15:57:19+03:00

Answer:

L=36 Inches

w=18 Inches

Step-by-step explanation:

Let:

w=Length of the Square ends of the package

l=Length of the package

Combined length and girth (perimeter of a cross section) =108 inches.

Girth +Length =108

4w+l=108

l=108-4w

Volume of the rectangular package, $V=w^2l$

Substituting l=108-4w

$V=w^2(108-4w)\\V(w)=108w^2-4w^3$

Now, we determine the critical points for V(w) and find out the dimensions which maximizes V.

The derivative of V

$V'(w)=216w-12w^2\\Setting \: V'(w)=0\\216w-12w^2=0\\216w=12w^2\\\text{Divide both sides by 12w}\\w=18 \: Inches$

Recall:

$l=108-4w=108-4(18)=36\: Inches$

The dimensions that will maximize the volume are:

L=36 Inches

w=18 Inches

Morgarella [4.7K] · Answer 2 · 2022-04-19T13:35:08+03:00

Answer:

The maximum volume of the package is obtained with a cross section of side 18 inches and a length of 36 inches.

Step-by-step explanation:

This is a optimization with restrictions problem.

The restriction is that the perimeter of the square cross section plus the length is equal to 108 inches (as we will maximize the volume, we wil use the maximum of length and cross section perimeter).

This restriction can be expressed as:

$4x+L=108$

being x: the side of the square of the cross section and L: length of the package.

The volume, that we want to maximize, is:

$V=x^2L$

If we express L in function of x using the restriction equation, we get:

$4x+L=108\\\\L=108-4x$

We replace L in the volume formula and we get

$V=x^2L=x^2*(108-4x)=-4x^3+108x^2$

To maximize the volume we derive and equal to 0

$\dfrac{dV}{dx}=-4*3x^2+108*2x=0\\\\\\-12x^2+216x=0\\\\-12x+216=0\\\\12x=216\\\\x=216/12=18$

We can replace x to calculate L:

$L=108-4x=108-4*18=108-72=36$

The maximum volume of the package is obtained with a cross section of side 18 inches and a length of 36 inches.