$y_1=ax+b$
is tangent to the curve $y_2=12x^2,$
with $0<x<\sqrt{12}.$
Find the reals $a,b$
such that the area delimited by $y_1,$
the $x$
axis and the $y$
axis is minimized.

 Find the equation of the line that is tangent to the curve
$y=(x1)^3$
at$x=0.$
 A rectangular cuboid (rectangular prism) has sides of length
$x,$
$x+2$
and$5x.$
Find the value of$x$
that maximizes its volume (make sure it gives a maximum).
 Find the equation of the line that is tangent to the curve

Try expressing the coordinates of the triangle vertices with respect to a single (and appropriate) variable.

For example, try to express the intersection points of the line with the
$x$
and$y$
axes with respect to the tangent point$x$
coordinate, call it$t$
? 
What is the area of the triangle in terms of those intersection points?

What about in terms of
$t$
? 
How does one find the maximum of a function of a single variable?

$y_1$
crosses the$x$
and$y$
axes at$y=b$
and$x=\frac{b}{a}$
respectively, and so the area in question (right triangle) is$A=\frac{b^2}{2a}.$
However, both$a$
and$b$
are functions of the$x$
coordinate of the tangent point, which we’ll denote with$t,$
i.e.$a=a(t)$
and$b=b(t),$
and hence$A=A(t).$
We must minimize the latter.At the tangent point
$x=t$
we equate the gradients (derivatives), and the values of the line and of the curve. Hence$a=2t$
and$y_1(t)=y_2(t),$
i.e.$b=12+t^2.$
Writing$\frac{dA}{dt} = 0$
we have:$\frac{d}{dt} \frac{(12+t^2)^2}{4t} =\frac{2(12+t^2)(2t)}{4t}\frac{(12+t^2)^2}{4t^2} = 0,$
which gives$4t^2=12+t^2,$
thus$t=2$
and$y_1=4x+16.$