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Determine the dimensions of a rectangular solid (with a square base) with maximum volume if its surface area is 529 square meters. Round each dimension to 4 decimal places

Sagot :

okpalawalter8

Answer:

[tex]x=23/\sqrt{6}[/tex],

[tex]y= \frac{23}{6\sqrt{6}}[/tex]

Step-by-step explanation:

From the question we are told that

Surface area is 529 square meters

Generally the dimensions of the cuboid

[tex]v=x^y[/tex]

[tex]D=4xy+2x^2 =529[/tex]

[tex]V=x^29(\frac{529-6x^2}{4x})[/tex]

[tex]V' =529-6x^2/4[/tex]

[tex]v'=0[/tex]

Therefore

[tex]529=6x^2[/tex]

[tex]x=\sqrt{\frac{529}{6} }[/tex]

[tex]x=23/\sqrt{6}[/tex]

Mathematically

[tex]v''=\frac{12x}{4}[/tex]

[tex]v''=-3\frac{23}{\sqrt{6} }<0[/tex]

[tex]y= \frac{529-2*(\frac{529}{6} }{4*\frac{23}{\sqrt{6} } }[/tex]

[tex]y= \frac{23}{6\sqrt{6}}[/tex]

Therefore

[tex]x=23/\sqrt{6}[/tex]

and

[tex]y= \frac{23}{6\sqrt{6}}[/tex]

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