. Applications in Optimization Optimization is a process of finding an extreme value (either maximum or minimum) under certain conditions. A procedure for solving for an extremum or optimization problems. Step : Draw an appropriate figure and label the quantities relevant to the problem.
Step : Find a experssion for the quantity to be maximized or minimized. Step : Using the given conditions of the problem, the quantity to be extremized . Step : Determine the interval of possible values for this variable from the conditions given in the problem. Step : Using the techniques of extremum (absolute extrimum, first derivative test or second derivative test) obtain the maximum or minimum.
Example . We have a unit square piece of thin material and want to make an open box by cutting small squares from the corners of our material and folding the sides up. The question is, which cut produces the box of maximum volume? Applications of Differential Calculus Let x = length of the cut on each side of the little squares.
V = the volume of the folded box. The length of the base after two cuts along each edge of size x is − x . The depth of the box after folding is x , so the volume is V × . Note that, when x = or , the volume is zero and hence there cannot be a box.
Therefore the problem is to maximize, V × ) , Now, dV dx = ( = ( )( x . dV dx = gives the stationary numbers x = , . Since ∉ ( , ) the only stationary number is at x = ( , ) . Further, dV dx changes its sign from postive to negative when passing through x = .
Therefore at x = the volume V is local maximum. The local maximum volume value is V = units. Hence the maximum cut can only be units. Example .
Find the points on the unit circle x = nearest and farthest from ( , ) . The distance from the point ( , ) to any point ( , ) x y is d . Instead of extremising d , for convenience we extremise D d ) , subject to the condition = . Now, dD −× , where the dy dx will be computed by differentiating = with respect to x .
Therefore, we get y dy which gives us dy =− Substituting this, we get dD [ ] dD dx = = ⇒ x Since ( , ) x y lie on the circle x = , we get x = which gives x = ± . Hence the points at which the extremum distance occur are, − . Fig. .