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Question: Two gas companies X and Y, where X is situated at (40,0) and Y is situated at (0,30), offer to insta...

Two gas companies X and Y, where X is situated at (40,0) and Y is situated at (0,30), offer to install equally priced gas furnaces in buyer s house. Company X adds a charge of Rs.40 per Km. of distance (measured along a straight line) between its location and buyers house, while company Y charges Rs.60 per Km. Of distance measured in the same way. Then the region where it is cheaper to have the furnace installed by the company X is:

A

The inside of the circle (x−54)2+(y+30)2=3600(x - 54)^2 + (y+30)^2 = 3600

B

The outside of the circle radius of circle is x2+y2+64x−108y−340≥0x^2 + y^2 + 64x-108y-340 \ge 0

C

Outside the circle the radius of circle is (x+32)2+(y−54)2=3600(x+32)^2 + (y - 54)^2 = 3600

D

Outside the circle (x+24)2+(y−12)2=2500(x + 24)^2 + (y -12)^2 = 2500

Answer

Outside the circle the radius of circle is (x+32)2+(y−54)2=3600(x+32)^2 + (y - 54)^2 = 3600

Explanation

Solution

The problem requires finding the region where the total cost from Company X is less than the total cost from Company Y. Let the buyer's house be at P(x,y)P(x, y). Company X is at X0(40,0)X_0(40, 0) with a charge of Rs. 40/Km, and Company Y is at Y0(0,30)Y_0(0, 30) with a charge of Rs. 60/Km.

The distance from the buyer's house to Company X is dX=(x−40)2+y2d_X = \sqrt{(x-40)^2 + y^2}. The cost for Company X is CX=40dX=40(x−40)2+y2C_X = 40 d_X = 40 \sqrt{(x-40)^2 + y^2}.

The distance from the buyer's house to Company Y is dY=x2+(y−30)2d_Y = \sqrt{x^2 + (y-30)^2}. The cost for Company Y is CY=60dY=60x2+(y−30)2C_Y = 60 d_Y = 60 \sqrt{x^2 + (y-30)^2}.

We want to find the region where CX<CYC_X < C_Y: 40(x−40)2+y2<60x2+(y−30)240 \sqrt{(x-40)^2 + y^2} < 60 \sqrt{x^2 + (y-30)^2}

Divide by 20: 2(x−40)2+y2<3x2+(y−30)22 \sqrt{(x-40)^2 + y^2} < 3 \sqrt{x^2 + (y-30)^2}

Square both sides: 4((x−40)2+y2)<9(x2+(y−30)2)4 ((x-40)^2 + y^2) < 9 (x^2 + (y-30)^2) 4(x2−80x+1600+y2)<9(x2+y2−60y+900)4 (x^2 - 80x + 1600 + y^2) < 9 (x^2 + y^2 - 60y + 900) 4x2−320x+6400+4y2<9x2+9y2−540y+81004x^2 - 320x + 6400 + 4y^2 < 9x^2 + 9y^2 - 540y + 8100

Rearrange the terms: 0<5x2+5y2+320x−540y+17000 < 5x^2 + 5y^2 + 320x - 540y + 1700

Divide by 5: 0<x2+y2+64x−108y+3400 < x^2 + y^2 + 64x - 108y + 340

To identify the region, we complete the square: (x2+64x+322)+(y2−108y+542)+340−322−542>0(x^2 + 64x + 32^2) + (y^2 - 108y + 54^2) + 340 - 32^2 - 54^2 > 0 (x+32)2+(y−54)2+340−1024−2916>0(x + 32)^2 + (y - 54)^2 + 340 - 1024 - 2916 > 0 (x+32)2+(y−54)2−3600>0(x + 32)^2 + (y - 54)^2 - 3600 > 0 (x+32)2+(y−54)2>3600(x + 32)^2 + (y - 54)^2 > 3600

This inequality represents the region outside the circle with center (−32,54)(-32, 54) and radius 3600=60\sqrt{3600} = 60. Option (C) correctly states this region.