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(B) Let the height and breadth of the sheet be $y \ m$ and $x \ m$ respectively.Given area of the sheet is $18 \ m^2$,so $x y = 18$.Converting margins

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the height and breadth of the sheet be $y \ m$ and $x \ m$ respectively.Given area of the sheet is $18 \ m^2$,so $x y = 18$.Converting margins to meters: top/bottom margins are $0.75 \ m$ each,side margins are $0.5 \ m$ each.The dimensions of the printable area are $(y - 1.5) \ m$ and $(x - 1) \ m$.The area available for printing is $A = (y - 1.5)(x - 1)$.Since $y = \frac{18}{x}$,we have $A = (\frac{18}{x} - 1.5)(x - 1) = 18 - \frac{18}{x} - 1.5x + 1.5 = 19.5 - \frac{18}{x} - 1.5x$.To maximize $A$,differentiate with respect to $x$: $\frac{dA}{dx} = \frac{18}{x^2} - 1.5$.Setting $\frac{dA}{dx} = 0$,we get $\frac{18}{x^2} = 1.5 \Rightarrow x^2 = \frac{18}{1.5} = 12$.Thus,$x = \sqrt{12} = 2 \sqrt{3} \ m$.Then $y = \frac{18}{2 \sqrt{3}} = \frac{9}{\sqrt{3}} = 3 \sqrt{3} \ m$.Checking the second derivative: $\frac{d^2A}{dx^2} = -\frac{36}{x^3}$,which is negative at $x = 2 \sqrt{3}$,confirming a maximum.Therefore,the height is $3 \sqrt{3} \ m$ and the breadth is $2 \sqrt{3} \ m$.

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