Question 855600
The way I see it, this is a calculus problem.
With {{{t}}}= time (in minutes) counted from the moment the base of the triangle is 2 feet and the height is 3 feet,
the base (in inches), height (in inches), and area (in square inches) are all functions of {{{t}}} :
{{{base=2*12-2t=24-2t}}}
{{{height=3*12+4t=36+4t}}}
{{{Area=base*height/2=(24-2t)(36+4t)/2=(864-72t+96t-8t^2)/2=(864+24t-8t^2)/2=432+12t-4t^2}}}
The rate of change of the area (in square inches per minute) is
{{{dArea/dt=12-8t}}} which changes with time.
At {{{t=0}}} (the moment the base of the triangle is 2 feet and the height is 3 feet)
{{{dArea/dt=12-8*1=highlight(12)}}}
 
Without invoking calculus, the only solution I see is saying that the rate is the slope of the tangent to the graph.
With {{{x=t}}} and {{{y=Area-432=12t-4t^2=12x-4x^2}}} the graph would be
{{{graph(300,300,-5,5,-25,25,12x-4x^2,12x)}}} .
The tangent {{{y=kx}}} , with slope {{{k}}} passes through the origin
and intersects {{{y=12x-4x^2}}} at only one point.
That means that {{{kx=12x-4x^2}}} must have only one solution.
{{{kx=12x-4x^2}}}<-->{{{4x^2-12x+kx=0}}}<-->{{{4x^2+(k-12)x=0}}}<-->{{{x(4x+k-12)=0}}}
has the solutions {{{x=0}}} and {{{x=(12-k)/4}}} .
Those solutions are the same only when {{{12-k=0}}}<-->{{{k=12}}} ,
which makes the line tangent.
A line with another slope would intersect the graph at two different points, with {{{x=0}}} , and {{{x=(12-k)/4}}} .