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

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.

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

Commented by ajfour last updated on 27/Jun/18

Right sketch,right answer Sir. Find area bounded by min{∣x∣,∣y∣}=2 max{∣x∣,∣y∣}=4

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

Commented by MrW3 last updated on 27/Jun/18

Commented by MrW3 last updated on 27/Jun/18

common area = 2

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

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=8^2 −4^2 =64−16=48

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

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

area formed by ∣x∣=4 and∣y∣=4 is8×8=64 area formed by ∣x∣=2 and ∣y∣=2 is=4×4=16 so=area required 64−16=48

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

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

Commented by ajfour last updated on 27/Jun/18

min{ } , max{ } has a meaning.

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

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

you solve the question...

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

Commented by MrW3 last updated on 27/Jun/18

area bounded by min{[x],[y]}=2 max{[x],[y]}=4 is 8. area bounded by min{∣x∣,∣y∣}=2 max{∣x∣,∣y∣}=4 is 16.

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

Commented by ajfour last updated on 27/Jun/18

Area for the second one is 16.

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

Commented by MrW3 last updated on 27/Jun/18

you′r right.

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

Commented by MrW3 last updated on 27/Jun/18

Commented by ajfour last updated on 27/Jun/18

Great, Sir. you are fast.

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

Commented by ajfour last updated on 28/Jun/18

Too Good Sir.

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

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 64−16=48.

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

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