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(B) Let height and breadth of the sheet be $y \ m$ and $x \ m$ respectively. Since the area is $18 \ m^2$,we have $x y = 18$. Converting to $cm$,$x y

Vedclass · Accepted Answer

$3 \sqrt{3} \ m, 2 \sqrt{3} \ m$

Vedclass · Answer

(B) Let height and breadth of the sheet be $y \ m$ and $x \ m$ respectively. Since the area is $18 \ m^2$,we have $x y = 18$. Converting to $cm$,$x y = 180000 \ cm^2$,so $y = \frac{180000}{x}$. The margins are $75 \ cm$ at top and bottom (total $150 \ cm = 1.5 \ m$) and $50 \ cm$ at each side (total $100 \ cm = 1 \ m$). The area available for printing is $A = (y - 1.5)(x - 1)$. Substituting $y = \frac{18}{x}$,we get $A = (\frac{18}{x} - 1.5)(x - 1) = 18 - \frac{18}{x} - 1.5x + 1.5 = 19.5 - 1.5x - \frac{18}{x}$. To maximize $A$,find $\frac{dA}{dx} = -1.5 + \frac{18}{x^2} = 0$. $\frac{18}{x^2} = 1.5 \Rightarrow x^2 = \frac{18}{1.5} = 12$. $x = \sqrt{12} = 2 \sqrt{3} \ m$. Then $y = \frac{18}{2 \sqrt{3}} = \frac{9}{\sqrt{3}} = 3 \sqrt{3} \ m$. Thus,the dimensions are height $3 \sqrt{3} \ m$ and breadth $2 \sqrt{3} \ m$.

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