A 100-inch piece of wire is divided into two pieces and each piece is bent into

Question

A 100-inch piece of wire is divided into two pieces and each piece is bent into a square. How should this be done in order to minimize the sum of the areas of the two squares? Express the sum of the areas of the squares in terms of the lengths x and y of the two pieces. What is the constraint equation relating x and y? Does this problem require optimization over an open interval or a closed interval? Solve the optimization problem. The legs of n right triangle have lengths a and b satisfying a + b= 10. Which values of a and b maximize the area of the triangle?

Explanation / Answer

1. Length of the two pieces are x and y inch.

clearly x+y = 100 which is the constraint related to x and y.

and the sum of the areas of the two square = A = x^2 + y^2

= x^2 + (100-x)^2

we minimize this sum with respect to x i.e. we calculate dA/dx and equate it to 0.

dA/dx = 2x - 2(100-x) = 4x-200

So dA/dx = 0 => 4x-200 = 0 => x = 50

2. the legs of a right angle triangle have length a and b unit.

a+b = 10

area of the triangle = (1/2)*ab sq unit.

Applying A.M.>= G.M. inequality on a and b, we get..

sqrt(ab) <= (a+b)/2 => ab <= [(a+b)/2]^2

=> (1/2)*ab <= [(a+b)^2]/8 = 100/8 = 12.5

maximum area will be 12.5 sq unit and it will be maximum when A.M.>= G.M. inequality attains equality. And it's equality condition is a = b.

So a = b = 5 will maximize the area. (Ans.)

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