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Consider an isosceles triangle whose two equal sides are of length 4 What is the largest possible area for such a triangle?

An isosceles triangle with side length 4 has an altitude x. By the Pythagorean theorem, the base of the triangle is 2*SQRT(16-x2). The area of the triangle is 1/2 base times height, so A=x*(16-x2)1/2. the derivative, dA/dx=(16-x2)1/2 - x2/(16-x2)1/2. This is found with the product rule and chain rule. This shows the rate which the area of the triangle changes with respect to the altitute. At the x value of the maximum, the area will have stopped increasing and begun to decrease, so the rate of increase wil be zero. We just need to solve for x. (16-x2)1/2 - x2/(16-x2)1/2=0 (16-x2)1/2=x2/(16-x2)1/2 (16-x2)=x2 16=2x2 8=x2 SQRT(8)=x. Now we can solve the original equation for the maximum are. SQRT(8)*SQRT(16-8) SQRT(8)*SQRT(8)=8 So 8 is the largest possible area.