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.
$$A = \frac{1}{4}(b + a) \sqrt{4a^2 - (b - a)^2}$$

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

 

Where $u = b + a$ and $v = \sqrt{4a^2 - (b - a)^2}$ from which $\dfrac{du}{db} = 1$. For $dv$ however, we will use the formula $d\left( \sqrt{u} \right) = \dfrac{du}{2\sqrt{u}}$. Hence, $\dfrac{dv}{db} = \dfrac{-2(b - a)}{2\sqrt{4a^2 - (b - a)^2}}$

Now, apply the whole $d(uv)$ to the equation:
$$\dfrac{dA}{db} = \dfrac{1}{4} \left[ (b + a) \cdot \dfrac{-2(b - a)}{2\sqrt{4a^2 - (b - a)^2}} + \sqrt{4a^2 - (b - a)^2} \cdot 1 \right] = 0$$