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MHT CET · Maths · Application of Derivatives

A poster is to be printed on a rectangular sheet of paper of area \(18 \mathrm{~m}^2\). The margins at the top and bottom of 75 cm each and at the sides 50 cm each are to be left. Then the dimensions i.e. height and breadth of the sheet so that the space available for printing is maximum, are _________ respectively.

  1. A \(2 \sqrt{3} \mathrm{~m}, 3 \sqrt{3} \mathrm{~m}\)
  2. B \(3 \sqrt{3} \mathrm{~m}, 2 \sqrt{3} \mathrm{~m}\)
  3. C \(3 \mathrm{~m}, 6 \mathrm{~m}\)
  4. D \(6 \mathrm{~m}, 3 \mathrm{~m}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3 \sqrt{3} \mathrm{~m}, 2 \sqrt{3} \mathrm{~m}\)

Step-by-step Solution

Detailed explanation


Let height and breadth of the sheet be ' \(y\) ' m and ' \(x\) ' m respectively.
\(\begin{array}{ll}
\therefore & x y=180000 \mathrm{~cm}^2 \\
\therefore & y=\frac{180000}{x}
\end{array}\)
\(\therefore \quad\) The area available for printing is
\(\begin{aligned}
A & =(y-150)(x-100) \\
& =\left(\frac{180000}{x}-150\right)(x-100) \\
& =180000-\frac{18000000}{x}-150 x-15000 \\
& =165000-150 x-\frac{18000000}{x}
\end{aligned}\)
\(\begin{array}{ll}
\therefore & \frac{\mathrm{dA}}{\mathrm{~d} x}=0-150+\frac{18000000}{x^2} \\
\therefore & \frac{\mathrm{dA}}{\mathrm{~d} x}=0 \Rightarrow x^2=\frac{18000000}{150}=120000 \\
& \Rightarrow x=200 \sqrt{3} \mathrm{~cm} \\
& \Rightarrow y=\frac{180000}{200 \sqrt{3}}=300 \sqrt{3} \mathrm{~cm}
\end{array}\)
Now, \(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{~d} x^2}=\frac{-36000000}{x^3}\)
\(\therefore \quad\) At \(x=200 \sqrt{3} \mathrm{~cm}, \frac{\mathrm{~d}^2 \mathrm{~A}}{\mathrm{~d} x^2} \lt 0\)
\(\therefore \quad\) Area is maximum at \(x=200 \sqrt{3} \mathrm{~cm}\) and \(y=300 \sqrt{3} \mathrm{~cm}\)
\(\therefore \quad y=3 \sqrt{3} \mathrm{~m}\) and \(x=2 \sqrt{3} \mathrm{~m}\)