Question

$\underline\mathtt{Question}$

A rectangular sheet of paper has its area 24 square metres the margin at the top and the bottom are 75 cm each and at the sides are 50 cm each what are the dimensions of the paper in the area of the printed space is maximum

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Answer 1

Answer:

here is your answer................

Answer 2

❍ $\Large{\underline{\underline{\sf{Required\:Answer:}}}}$

• Firstly convert the length and breadth of margins into metres. For that multiply them with $\displaystyle{\bf{\dfrac{1}{100}}}$

$\implies{\sf{50\times \dfrac{1}{100} = 0.5}}$

$\implies{\sf{75\times \dfrac{1}{100} = 0.75}}$

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Let's assume that length(l) = x.

Then the breadth(b) will be = $\displaystyle{\bf{\dfrac{24}{x}}}$

And A be the area of the printed space.

Then,

the dimensions of the printed area = $\displaystyle{\sf{\bigg[x-(0.75+0.75)\bigg] and \bigg[ \dfrac{24}{x} - (0.5+0.5)\bigg]}}$

$\large{\star}{\boxed{\sf{\orange{Area=length\times breadth}}}}{\star}$

• Substitute the values

$\implies{\sf{Area = \bigg[x-(0.75+0.75)\bigg] \times \bigg[ \dfrac{24}{x} - (0.5+0.5)\bigg]}}$

$\implies{\sf{Area = (x-1.5) \times \bigg( \dfrac{24}{x} - 1\bigg)}}$

$\implies{\sf{Area = 24 -x -1.5\bigg(\dfrac{24}{x}\bigg) -1.5}}$

$\implies{\sf{Area = 25.5 - 1.5x - \dfrac{24}{x}}}$

• Now, Area is maximum if $\displaystyle{\rm{ \dfrac{dA}{dx} = 0}}$
• Substitute the values.

$\displaystyle{\sf{\:\:\:\:\dfrac{dA}{dx}=25.5 - 1.5x - \dfrac{24}{x^{2}}}}$

:$\implies{\sf{0=25.5 - 1.5x - \dfrac{24}{x^{2}}}}$

:$\implies{\sf{0=-\dfrac{3}{2}+\dfrac{24}{x^{2}}}}$

:$\implies{\sf{\dfrac{3}{2}=\dfrac{24}{x^{2}}}}$

:$\implies{\sf{x^{2}=24\times \dfrac{2}{3} - 16}}$

Again, $\displaystyle{\sf{\dfrac{d^{2}A}{dx^{3}}=0+24(-2x^{-3}) = \dfrac{48}{x^{3}}}}$

$\implies{\sf{ \dfrac{d^{2}A}{dx^{3}}-\dfrac{48}{x^{3}}= \dfrac{-3}{4}}}$

$\displaystyle{\sf{\dfrac{-3}{4}<0}}$

Therefore, A or Area of printed space is maximum when

$\large{\boxed{\sf{\red{x\:or\:length = 4.}}}}$

• Substitute x = 4 in $\displaystyle{\rm{\dfrac{24}{x}}}$

$\implies{\sf{\dfrac{24}{x}}}$

$\implies{\sf{\dfrac{24}{4}=6}}$

$\large{\boxed{\sf{\red{ Breadth = 6.}}}}$

Hence, the area of printed space is maximum when length and breadth of the sheet is 4m and 6m respectively.

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