Find-the-area-common-to-min-x-y-2-and-max-x-y-4-x-denotes-the-greatest-integer-less-than-or-equal-to-x-

Question Number 38568 by ajfour last updated on 27/Jun/18

$${Find}\:{the}\:{area}\:{common}\:{to} \\ $$$${min}\left\{\left[{x}\right],\:\left[{y}\right]\:\right\}\:=\mathrm{2}\:\:\:{and} \\ $$$${max}\left\{\left[{x}\right],\:\left[{y}\right]\:\right\}\:=\mathrm{4}\:. \\ $$$$\left[{x}\right]\:{denotes}\:{the}\:{greatest}\:{integer} \\ $$$${less}\:{than}\:{or}\:{equal}\:{to}\:{x}. \\ $$

Commented by ajfour last updated on 27/Jun/18

$${Right}\:{sketch},{right}\:{answer}\:{Sir}. \\ $$$${Find}\:{area}\:{bounded}\:{by} \\ $$$${min}\left\{\mid{x}\mid,\mid{y}\mid\right\}=\mathrm{2} \\ $$$${max}\left\{\mid{x}\mid,\mid{y}\mid\right\}=\mathrm{4} \\ $$

Commented by MrW3 last updated on 27/Jun/18

Commented by MrW3 last updated on 27/Jun/18

$${common}\:{area}\:=\:\mathrm{2} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

$${area}=\mathrm{8}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} =\mathrm{64}−\mathrm{16}=\mathrm{48} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

$${area}\:{formed}\:{by}\:\mid{x}\mid=\mathrm{4}\:{and}\mid{y}\mid=\mathrm{4}\:\:{is}\mathrm{8}×\mathrm{8}=\mathrm{64} \\ $$$${area}\:{formed}\:{by}\:\mid{x}\mid=\mathrm{2}\:{and}\:\mid{y}\mid=\mathrm{2}\:{is}=\mathrm{4}×\mathrm{4}=\mathrm{16} \\ $$$${so}={area}\:{required}\:\mathrm{64}−\mathrm{16}=\mathrm{48} \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

Commented by ajfour last updated on 27/Jun/18

$${min}\left\{\:\right\}\:,\:{max}\left\{\:\right\}\:{has}\:{a}\:{meaning}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

$${you}\:{solve}\:{the}\:{question}… \\ $$

Commented by MrW3 last updated on 27/Jun/18

$${area}\:{bounded}\:{by} \\ $$$${min}\left\{\left[{x}\right],\left[{y}\right]\right\}=\mathrm{2} \\ $$$${max}\left\{\left[{x}\right],\left[{y}\right]\right\}=\mathrm{4} \\ $$$${is}\:\mathrm{8}. \\ $$$$ \\ $$$${area}\:{bounded}\:{by} \\ $$$${min}\left\{\mid{x}\mid,\mid{y}\mid\right\}=\mathrm{2} \\ $$$${max}\left\{\mid{x}\mid,\mid{y}\mid\right\}=\mathrm{4} \\ $$$${is}\:\mathrm{16}. \\ $$

Commented by ajfour last updated on 27/Jun/18

$${Area}\:{for}\:{the}\:{second}\:{one}\:{is}\:\mathrm{16}. \\ $$

Commented by MrW3 last updated on 27/Jun/18

$${you}'{r}\:{right}. \\ $$

Commented by MrW3 last updated on 27/Jun/18

Commented by ajfour last updated on 27/Jun/18

$${Great},\:\:{Sir}.\:{you}\:{are}\:{fast}. \\ $$

Commented by ajfour last updated on 28/Jun/18

$${Too}\:{Good}\:{Sir}. \\ $$

Commented by MrW3 last updated on 28/Jun/18

Commented by MrW3 last updated on 28/Jun/18

$${the}\:{pink}\:{color}\:{area}\:{is}\:{also}\:{bounded} \\ $$$${area}\:{which}\:{is}\:\mathrm{64}−\mathrm{16}=\mathrm{48}. \\ $$

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