Answer:
Maximum length and breadth of the box is 6.32 cm and the height of the box is 3.16 cm
Step-by-step explanation:
Surface area of box
[tex]4xy+x^2=120\\\Rightarrow y=\dfrac{120-x^2}{4x}[/tex]
Volume of box is
[tex]V=x^2y\\\Rightarrow V=x^2\times \dfrac{120-x^2}{4x}\\\Rightarrow V=\dfrac{120x-x^3}{4}\\\Rightarrow V=30x-\dfrac{1}{4}x^3[/tex]
Differentiating with respect to [tex]x[/tex]
[tex]\dfrac{dV}{dx}=30-\dfrac{3}{4}x^2[/tex]
Equating with zero
[tex]30-\dfrac{3}{4}x^2=0\\\Rightarrow x^2=\dfrac{30\times 4}{3}\\\Rightarrow x^2=40\\\Rightarrow x=6.32[/tex]
Double derivative of the volume
[tex]\dfrac{d^2V}{dx^2}=-\dfrac{6x}{4}=-\dfrac{6\times 6.32}{4}=-9.48<0[/tex]
So, the volume is maximum at [tex]x=6.32[/tex]
[tex]y=\dfrac{120-x^2}{4x}=\dfrac{120-40}{4\times \sqrt{40}}\\\Rightarrow y=3.16[/tex]
So, the maximum length and breadth of the box is 6.32 cm and the height of the box is 3.16 cm.
