Given $f\colon\Bbb{R}^2\to\Bbb{R}$ such that $f'(A,(a,b))=ab+3b+4a$ find maximum derivative

Question

Given $f\colon\Bbb{R}^2\to\Bbb{R}$ such that $$f'(A,(a,b))=ab+3b+4a$$ find the value of maximum derivative at point $A$, knowing that $f\in\mathcal{C}^1$.

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We observe that $$f'(A,(1,0))=f_x(A)=4,\qquad f'(A,(0,1))=f_y(A)=3,$$ so as $f\in\mathcal{C}^1$ we observe that $$\vec{\nabla}f(A)=(f_x(A),f_y(A))=(4,3).$$ We need to find $f'_{\max}(A)$, so $$f'_{\max}(A)=\lVert\vec{\nabla}f(A)\rVert=\sqrt{4^2+3^2}=5.$$ Is it correct?

P.S. Some of you asked what does $f(A,(a,b))$ mean. I suspect that this notation refers to Directional derivative of $f$ at point $A$ in direction $\vec{v}=(a,b)\in\Bbb{R}^2$.

Answer 1

I'm going to go against the grain and say no for the following pedantic reason: there does not exist a $C^1$ function $f\colon \mathbb R^2\to\mathbb R$ and a point $A\in\mathbb R^2$ satisfying $$ f'(A,(a,b))=ab+3b+4a,\qquad \forall (a,b)\in\mathbb R^2. $$ Indeed, this follows from the observation that $$ f'(A,(a,b))=af_x(A)+bf_y(A), $$ so there is no opportunity for an $ab$ term to enter.

Answer 2

As pre-kidney noted, a function with such directional derivative won't be $\mathcal C^1$. You can still try to solve the problem though.

If by $f_{\rm max}'(A)$ you mean $$f_{\rm max}'(A) = \sup_{v\in\mathbb R^2} \frac{|f'(A,v)|}{||v||} $$ then your method is not correct. The method works in the question that you linked in the comments, but won't work here because here the directional derivative is nonlinear.

You have \begin{align}\sup_{v\in\mathbb R^2} \frac{|f'(A,v)|}{||v||} &= \sup_{v\in\mathbb R^2,||v||=1} |f'(A,v)| = \sup_{(a,b): a^2+b^2=1} |f'(A,(a,b))| = \\ &= \sup_{(a,b): a^2+b^2=1} |ab+4a+3b|\end{align} Using the method of Lagrange multipliers, you can reduce the problem of finding the supremum to solving the system of equations: $$ \left\{\begin{array}{l} b + 4 + 2\lambda a = 0 \\ a + 3 + 2\lambda b = 0 \\ a^2+b^2 =1 \end{array}\right. $$ This unfortunately gives you $$ \left\{\begin{array}{l} a = \frac{-8\lambda +3}{4\lambda^2-1} \\ b = \frac{-6\lambda +4}{4\lambda^2-1} \\ (-6\lambda +4)^2 + (-8\lambda +3)^2 =(4\lambda^2-1)^2 \end{array}\right. $$ And that last equation doesn't seem to have exact solutions, you can only solve it numerically.