x x - - To find the extrema, we apply second derivative test. So, d D = The value of d D d D > ; This implies the nearest and farthest points are and respectively. Therefore, the nearest and the farthest distances are respectively - and + . Example .
A steel plant is capable of producing x tonnes per day of a low-grade steel and y tonnes per day of a high-grade steel, where y . If the fixed market price of low-grade steel is half that of high-grade steel, then what should be optimal productions in low-grade steel and high-grade steel in order to have maximum receipts. Let the price of low-grade steel be ` p per tonne. Then the price of high-grade steel is ` p per tonne.
The total receipt per day is given by R px py px . Hence the problem is to maximise R . Now, simplifying and differentiating R with respect to x , we get R = p dR dx = p x d R = - - Now, dR dx = ⇒ and hence x = ± At x d R < and hence R will be maximum. If x = - then y = .
Therefore the steel plant must produce low-grade and high-grade steels respectively in tonnes per day are - and Example . Prove that among all the rectangles of the given area square has the least perimeter. Let x y be the sides of the rectangle. Hence the area of the rectangle is xy k (given).
The perimeter of the rectangle P is ( . So the problem is to minimize ( suject to the condition k . Let P x k ( ) = - - Applications of Differential Calculus