A poster is to be printed on a rectangular sheet of paper of area $18 \mathrm{~m}^2$. The margins at the top…

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 $75 \mathrm{~cm}$ each and at the sides $50 \mathrm{~cm}$ 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. $2 \sqrt{3} \mathrm{~m}, 3 \sqrt{3} \mathrm{~m}$
  2. $3 \sqrt{3} \mathrm{~m}, 2 \sqrt{3} \mathrm{~m}$
  3. $3 \mathrm{~m}, 6 \mathrm{~m}$
  4. $6 \mathrm{~m}, 3 \mathrm{~m}$

Solution

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

Asked in: MHT CET 2023 (12 May Shift 2)

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