Hello Angelina Lopez,
1.) We are told from the given information the following:
$$f(3)=4$$
$$f'(3)=0$$
This will give us two linear equations in two unknowns:
$$f(3)=a(3-1)^2+b(3-2)^2=4a+b=4$$
$$f'(3)=2a(3-1)+2b(3-2)=4a+2b=0$$
Subtracting the first equation from the second, we eliminate $a$ to obtain:
$$b=-4\implies a=2$$
We may use the second derivative test to determine the nature of the extremum:
$$f''(3)=2(2)+2(-4)=-4$$
Thus, the extremum is a maximum, and we know it is a global maximum since the second derivative is a constant, because the function is quadratic.
2.) Let $a$ and $b$ be the legs and $c$ be the hypotenuse. Since the triangle is a right triangle, we have by Pythagoras:
$$a^2+b^2=c^2$$
Implicitly differentiating with respect to time $t$, and dividing through by $2$, we obtain:
$$a\frac{da}{dt}+ b\frac{db}{dt}= c\frac{dc}{dt}$$
If $b$ is the vertical leg, then we know:
$$\frac{db}{dt}=0$$
and so we have:
$$a\frac{da}{dt}=c\frac{dc}{dt}$$
or:
(1) $$\frac{dc}{dt}=\frac{a}{\sqrt{a^2+b^2}}\frac{da}{dt}$$
Now, the area $A$ of the triangle is:
$$A=\frac{1}{2}ab$$
Implicitly differentiating with respect to time $t$, we find:
$$\frac{dA}{dt}=\frac{1}{2} \left(a\frac{db}{dt}+ \frac{da}{dt}b \right)$$
Using $$\frac{db}{dt}=0$$ this becomes:
$$\frac{dA}{dt}=\frac{b}{2}\frac{da}{dt}$$
or:
$$\frac{da}{dt}=\frac{2}{b}\frac{dA}{dt}$$
Substituting this into (1), we obtain:
$$\frac{dc}{dt}=\frac{a}{\sqrt{a^2+b^2}} \frac{2}{b}\frac{dA}{dt}$$
$$\frac{dc}{dt}=\frac{2a}{b\sqrt{a^2+b^2}}\frac{dA}{dt}$$
Using the given data:
$$a=12\text{ cm},\,b=5\text{ cm},\,\frac{dA}{dt}=10\,\frac{\text{cm}^2}{\text{s}}=\frac{48}{13}\,\frac{\text{cm}}{\text{s}}$$
we then find:
$$\frac{dc}{dt}=\frac{2\left(12\text{ cm} \right)}{\left(5\text{ cm} \right)\sqrt{\left(12\text{ cm} \right)^2+\left(5\text{ cm} \right)^2}}\left(10\,\frac{\text{cm}^2}{\text{s}} \right)=\frac{48}{13}\,\frac{\text{cm}}{\text{s}}$$