First we draw a picture. Here is a rectangle with an area of {{MATH_62}}. {{CMD_63}} {{ENV_64}} {{CMD_65}} The perimeter of the rectangle is {{MATH_66}} In order to be able to use calculus techniques, we have to express the perimeter as a function of a single variable. To this end, we have to express the variable, say {{MATH_67}}, in terms of {{MATH_68}}, using the constraint, {{MATH_69}}, {{MATH_70}} Hence the perimeter of a rectangle with area {{MATH_71}} can be expressed as a function of a single variable, {{MATH_72}}, {{MATH_73}} Again, for the side lengths to be physically relevant, we must assume that {{MATH_74}}. Therefore, the domain of the function {{MATH_75}} is the interval {{MATH_76}}. So, to minimize the perimeter of the rectangle, we have to find the global minimum of the function {{MATH_77}} on the interval {{MATH_78}}. Since the domain of {{MATH_79}} is an open interval, the global minimum (if it exists!) has to be a local minimum and, therefore, it occurs at a critical point. So, we have to find all critical points of {{MATH_80}} on the interval {{MATH_81}}. Since {{MATH_82}} in order to find the critical points, we have to solve the equation {{MATH_83}} Solving for {{MATH_84}} gives us {{MATH_85}}. Only the value {{MATH_86}} is in the domain of {{MATH_87}}. Since {{MATH_88}} is defined everywhere on the interval {{MATH_89}}, there are no more critical values, and there are no endpoints. Is there a local maximum, minimum, or neither at {{MATH_90}}? The second derivative is {{MATH_91}} and {{MATH_92}}, so there is a local minimum. Since there is only one critical point, this is also the global minimum.