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
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
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.
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