5. Let for and f(x)=12X^2 for x>=0 and f(x)>=0

Question

a. The line tangent to the graph of f at the point (k, f(k)) intercepts the x-axis at x = 4. What is the value of k?
b. An isosceles triangle whose base is the interval from (0, 0) to (c, 0) has its vertex on the graph of f. For what value of c does the triangle have a maximum area? Justify your answer.

Answer 1

dy/dx = 24x
at (k, 12k^2)
dy/dx = 24k

equation of tangent:
y - 12k^2 = 24k(x-k)
but the point (4,0) is on this line
0-12k^2 = 24k(4-k)
-12k^2 = 96k - 24k^2
12k^2 - 96k = 0
k(k-8) = 0
k = 0 or k = 8

at k=0, the tangent would be the x-axis itself, which does pass through the point (4,0)
More than likely they want k=8

check:
if k=8 , the point on the curve is (8, 768)
slope of tangent is 96
tangent equation at (8,768) is
y - 768 = 96(x-8)
for x-intercept, let y = 0
-768 = 96x - 768
0 = 96x
x = 0

Answer 2

If the base is form (0,0) to (c,0) and the vertex of the isosceles triangle lies on y = 12x^2, then the altitude must right-bisect the base
So if we let the vertex be (a,12a^2) then the base will be on (0,0 and (2a,0)
Area = (1/2)(2a)(12a^2) = 12a^3

As a gets larger, 12a^3 gets larger and there is no maximum area.