Maxima and minima (trapezoidal gutter)

Submitted by perfectveneer777 on

I have a problem sir.

h=1/2√4a^2-(b-a)^2

A=1/4(b+a)√4a^2-(b-a)^2

dA/db=1/4[(b+a)-2(b-a)/2√4a^2-(b-a)^2 + √4a^2-(b-a)^2]=0.

Question:
How did it become +√4a^2-(b-a)^2]=0?
How did it become zero? Why is there zero?

Tags
Maxima and minima

Jhun Vert Wed, 07/24/2024 - 09:53

Based on the differentiation dA/db, It can be seen that a is constant from the equation.
A=14(b+a)√4a2−(b−a)2

Use the product formula of differentiation
d(uv)=u dv+v du

 

Where u=b+a and v=√4a2−(b−a)2 from which dudb=1. For dv however, we will use the formula d(√u)=du2√u. Hence, dvdb=−2(b−a)2√4a2−(b−a)2

Now, apply the whole d(uv) to the equation:
dAdb=14[(b+a)⋅−2(b−a)2√4a2−(b−a)2+√4a2−(b−a)2⋅1]=0

The zero is the concept of maxima and minima. You need to go back to the basic concept of optimization to understand why the equation is equated to zero.