CALC: Find the dimensions of a rectangle with perimeter 68 m whose area is as large as possible…?

2 Answers

• Let the sides of the rectangle be length x and y and the area be A

  Then 2x+2y=68

  So x+y=34

  y=34-x

  A=xy

  Substitute the value of y into the equation for A

  A=x(34-x)

  A=34x-x^2

  dA/dx=34-2x

  Maximum occurs at dA/dx=0

  So 34 - 2x = 0

  2x=34

  x=17

  since x+y = 34, y=34

  So the largest rectangle with perimeter 68m is the square with sides of length 17m

• certainly, any rectangle will artwork. If it would not artwork before each thing, only exchange the units you're utilising to degree. as an occasion, think of of a rectangle 12 inches by utilising 18 inches. It has a edge of 60 inches and component to 216 sq. inches, so curiously that the fringe is below the section. whether, if we alter units and degree in ft, then the rectangle is one million foot by utilising one million.5 ft, so it has a edge of 5 ft, and and component to one million.5 sq. ft. Now curiously that the fringe is larger than the section. you're able to do the comparable with any rectangle. in case you utilize units that are sufficiently enormous, the fringe will finally end up bigger than the section. This unusual situation is why, as persons have suggested, we actually shouldn't attempt to evaluate measurements of diverse categories, like length, section, and volume.