Answer to Question #291522 in Calculus for Kang

Question

Postal regulations require that a parcel post package be no greater than 3m in the sum of its length and perimeter of its cross-section (girth). What is the volume in cubic meter 9f the largest package allowed by the postal regulations if the package is to be rectangular in shape and has square ends?

Answer 1

Solution:

$3\ m=300\ cm$

Let the length be $L\ cm$ and width & height be $x\ cm$ each.

Volume, $V=L\times x\times x=Lx^2\ ...(i)$

Also, given that $4x+L=300$

$\Rightarrow L=300-4x\ ...(ii)$

Put (ii) in (i).

$V=(300-4x)x^2 \\\Rightarrow V=300x^2-4x^3 \\\Rightarrow V'=600x-12x^2$

Put V'=0

$\Rightarrow 600x-12x^2=0 \\ \Rightarrow 12x(50-x)=0 \\\Rightarrow x=0, x=50$

Reject $x=0$

Now, $V''=600-24x$

At $x=50, V''=600-24(50)=-600<0$

So, maxima exists.

Put $x=50$ in (ii)

$L=300-4(50)=100\ cm$

Then, $V=Lx^2=100(50)^2=25000\ cm^3$