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A corner lot has the form of a right triangle; it has perpendicular sides measuring 60 m and 80 m,
respectively. It is required to construct the largest rectangular building with sides parallel to the street.
What is the largest area of the building?

 Feb 3, 2021
 #1
avatar+113120 
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Have you done calculus yet?

 Feb 3, 2021
 #2
avatar+118470 
+2

See the following  image  :

 

 

Let  x be the width of such a  building

 

The segment  BC  has  a  slope  of   (-80) / (60)  =  -8/6  = -4/3

 

And the  equation  of  a line theorugh this segmnt has the  equation  y = (-4/3) ( x - 60)

 

Note  that at any point x,  the length  of the  building will  be  equal   to this "y"

 

So.....the  area of such a  building  is  x * y  =    x [ ( -4/3) (x -60) ]

 

So

 

Area   =    (-4/3) (x) ( x -60)  =    (-4/3)x^2  + 80x

 

This is a parabola that  turns  downward

 

The  "x"   that will maximize  the   area =  the x coordinate  of the  vertex  =     -80  / (2 *-4/3)   =  80 / (8/3)  = 240 / 8 =  30 m

 

And  the  max area is     (-4/3)(30)^2  + 80 (30)  =  -1200  +  2400  =   1200 m^2

 

{The building is denoted  by ADFE}

 

cool cool cool

 Feb 3, 2021
edited by CPhill  Feb 3, 2021

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