3 m=300 cm
Let the length be L cm and width & height be x cm each.
Volume, V=L×x×x=Lx2 ...(i)
Also, given that 4x+L=300
⇒L=300−4x ...(ii)
Put (ii) in (i).
V=(300−4x)x2⇒V=300x2−4x3⇒V′=600x−12x2
Put V'=0
⇒600x−12x2=0⇒12x(50−x)=0⇒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=Lx2=100(50)2=25000 cm3
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