A rectangle has a perimeter of 60m. Use Calculus to show that the maximum area of this rectangle is 225 square centimetres

Question

 Optimization 

Answer 1

To begin, we know that the perimeter P=2L+2W, where L is the length and W is the width. Then P=2L+2W. Since P=60, we have 60=2L+2W. We know that Area A=LW. Solve 60=2L+2W for L to get L=(60-2W)/2=30-W. Substituting back into our area equation, we have A=(30-W)(W)=30W-W². Taking the derivative of each side, we get dA=30-2W. The Area function is a parabola that opens down. To find the place at which we have maximum area, we want to find the maximum of the parabola which will occur when the derivative is 0 (the tangent line at the maximum point has slope 0). So we will set 30-2W=0 to find the critical points. Solving for W we get W=15. So the width is 15 when the Area is maximized. We can easily find L using 60=2L+2W. Plugging in W=15 we get 60=2L+2(15) which simplifies to L=15. Thus, the area A=15*15=225.