2 Maximizing rectangle perimeters All rectangles with an area of 64 m have a perimeter given ? by P(? ) = 2 x + 128/x. ?wher? is the length of one. side of the rectangle. Find the absolute minimum value of the perimeter function. What are the Dimensions of the rectangle with minimum perimeter?
STEP_BY_STEP SOLUTION Step-1 Let f be a continuous function defined on an open interval containing a 1 number ‘c’.The number ‘c’ is critical value ( or critical number ). If f (c) 1 = 0 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). Step-2 When an output value of a function is a maximum or a minimum over the entire domain of the function, the value is called the absolute maximum or the absolute minimum. Let f be a func tion with domain D and let c be a fixed constant in D . Then the output value f( ) is the 1. Absolute maximum value of f on D if and only if f(x) f(c) , for all x in D. 2. Absolute minimum value of f on D if and only if f(c) f(x) , for all x in D. Step_3 Given that ; x is the length of one side of the rectangle. Area of the rectangle is 64 m 2 128 Perimeter of the given rectangle is P(x) = 2x+ x .Clearly P(x) is a rational function and it is continuous for all x ,except zero. But here x is length of a rectangle , so it takes only positive values .Hence the function is continuous on this particular domain. Now , we have to find out the absolute minimum value of the perimeter function .For that we have to evaluate the function at critical points.so, the critical points satisfies the equation . Given function is; P(x) = 2x+ 128 . For the critical points we have to differentiate the x function with respect to n .[ Since from the step-1] d d 128 dx (P(x)) = dx ( 2x+ x ). 1 d d 128 d d P (x) = dx( 2x) + dx( x ) ,since dx(u+v) = dx(u)+ = 2 d ( ) + 128 d ( ),s ince d (Cx) = C dx dx x dx = 2(1) +128( 2 ), since d (x ) = n x n1 , x dx = 2 - 1x8 . 1 128 Since , from the definition f (c) = 0 = 2 - c 2 - 128 = 0 c2 That is , 128 = 2 64 = c 2 c Therefore , c = -8 , 8. Hence ,x = 8 lies in the domain , since x is the length of a rectangle it should be positive number. Therefore , we can say that P(x) has minimum perimeter value at x = 8m. 128 So, the minimum value of the perimeter is ; P(8) = (2(8) + 8 )m = (16 + 16)m = (32)m. Therefore , the minimum value of the perimeter is 32m.