A line $y_1=ax+b$ is tangent to the curve $y_2=12-x^2,$ 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.
1. Find the equation of the line that is tangent to the curve $y=(x-1)^3$ at $x=0.$
2. A rectangular cuboid (rectangular prism) has sides of length $x,$ $x+2$ and $5-x.$ Find the value of $x$ that maximizes its volume (make sure it gives a maximum).

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 $y_1=-4x+16.$