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find the minimum surface area of a rectangular box with a square base and top vertical sides and a volume of 27 cubic meters draw the pictures

User Scrubbie
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1 Answer

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The dimensions for the surface area are
\( x = 3 \)m and
\( h = 3 \)m.

How did we get the value?

Assume that the length of a side of the rectangular base and top is (x) and the length of the vertical side is (h). The volume of the box is given by the product of the base area and the height. Now, equate this volume with the given volume of 27 cubic meters.


\[ V = x^2 h \]

Given that the volume is 27 cubic meters:


\[ x^2 h = 27 \]

Express the surface area
\( A \) in terms of
\( x \) and
\( h \).

The surface area of the box is
\[ A = 2x^2 + 4xh \]

Substitute the expression for
\( h \) from the volume equation:


\[ A = 2x^2 + 4x \left( (27)/(x^2) \right) \]


\[ A = 2x^2 + (108)/(x) \]

Take the derivative of
\( A \) with respect to
\( x \) and set it equal to zero:


\[ (dA)/(dx) = 4x - (108)/(x^2) \\= 0 \]

Solving for
\( x \):


\[ 4x = (108)/(x^2) \]


\[ x^3 = 27 \]


\[ x = 3 \]

Getting the value of
\( x \), substitute it back into the volume equation to find
\( h \):


\[ 3^2 h = 27 \]


\[ 9h = 27 \]


\[ h = 3 \]

So, the dimensions for the minimum surface area are
\( x = 3 \) and
\( h = 3 \).

find the minimum surface area of a rectangular box with a square base and top vertical-example-1
User Jay Zelos
by
8.6k points

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