Answer to Question #213181 in Calculus for renz ash

Question

question #213175

Answer 1

Here is the answer of your ques. 213175

Total Volume of box V = $b(b+1)(b+4)=b(b^2+5b+4)=b^3+5b^2+4b$

Differentiate V w.r.t. b-

$\dfrac{dV}{db}=3b^2+10b+4$

Hence Volume is increasing at $3b^2+10b+4$