Question

A poster is to have an area of \({\bf{180}}\;{\bf{i}}{{\bf{n}}^{\bf{2}}}\) with 1 inch margins at the bottom and sides and a 2-inch margin at the top. What dimensions will given the largest printed area?

Short answer

The dimensions of the poster are \(2\sqrt {30} \;{\rm{in}}{\rm{.}}\) and \(3\sqrt {30} \;{\rm{in}}{\rm{.}}\).

Step-by-step solution

Find the function for the area of the window

The figure below represents the sketch of the poster.

The area of the rectangle inside is:

\(\begin{aligned}{c}xy = 180\\y = \frac{{180}}{x}\end{aligned}\)

The printed area is:

\(\begin{aligned}{c}A\left( x \right) = \left( {x - 2} \right)\left( {y - 3} \right)\\ = \left( {x - 2} \right)\left( {\frac{{180}}{x} - 3} \right)\\ = 186 - 3x - \frac{{360}}{x}\end{aligned}\)

Differentiate the function of the area of the poster

Differentiate the function \(A\left( x \right) = 186 - 3x - \frac{{360}}{x}\).

\(\begin{aligned}{c}A'\left( x \right) = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {186 - 3x - \frac{{360}}{x}} \right)\\ = 0 - 3 + \frac{{360}}{{{x^2}}}\\ = \frac{{360}}{{{x^2}}} - 3\end{aligned}\)

Differentiate the function \(A'\left( x \right)\).

\(\begin{aligned}{c}A''\left( x \right) = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{{360}}{{{x^2}}} - 3} \right)\\ = - \frac{{1080}}{{{x^3}}}\end{aligned}\)

Find the values of x at which total area of poster is maximum

Find the roots of \(A'\left( x \right) = 0\).

\(\begin{aligned}{c}\frac{{360}}{{{x^2}}} - 3 = 0\\{x^2} = 120\\x = 2\sqrt {30} \end{aligned}\)

As \(A''\left( x \right) < 0\)therefore the area of the window is maximum for \(x = 2\sqrt {30} \).

Find the value of y

Substitute \(2\sqrt {30} \) for x in the equation \(y = \frac{{180}}{x}\).

\(\begin{aligned}{c}y = \frac{{180}}{{2\sqrt {30} }}\\ = 3\sqrt {30} \end{aligned}\)

Thus, the dimensions of the poster are \(2\sqrt {30} \;{\rm{in}}.\) and \(3\sqrt {30} {\rm{ in}}{\rm{.}}\).