Find the dimensions and maximum area of a rectangle, if it’s perimeter is 24 inches

The length and the width of the rectangle are 6 inches and 6 inches respectively, and the maximum area of a rectangle is 36 square inches if its perimeter is 24 inches It is given that the perimeter of a rectangle is 24.It is required to find the dimensions and maximum area of a rectangle.It is a question about finding maxima. What are maxima and minima?Maxima and minima of a function are the extrema within the range, in other words, the maximum value of a function at a certain point is called maxima and the minimum value of a function at a certain point is called minima.We know the perimeter of a rectangle (p) = 2(l+w)and area of a rectangle (a) = l×wWhere l = length and w = widthWe have p = 2(l+w) = 24 ⇒ l+w = 12 ⇒ w = 12 - land         a = l×w or,           a = l(12--l)or,           [tex]\rm a = 12 l - l^2[/tex]To find the maxima and minima of any function we differentiate the function and equate it to zero.a' = 12- 2l a' = 0 ⇒ 12 - 2l = 0 ⇒ l = 6 inchSince a'' ( double derivative)  is less than zero ie.  [tex]\rm a^{''} \leq 0[/tex] hence the maximum area of the rectangle will be at l = 6 inch.⇒ w = 12 - l ⇒ 12 - 6 = 6 inchand the maximum area will be,a = l × w = 6 ×6 = 36 inch sqaure.Thus, the length and the width of the rectangle are 6 inches and 6 inches respectively, and the maximum area of a rectangle is 36 square inches if its perimeter is 24 inches Know more about the maxima and minima here:brainly.com/question/6422517