If y=3x-7, x is greater than 0, what is the minimum product of x^2y?

Question

A. -5.646
B. 0
C. 1.556
D. 2.813
E. 4.841

Answer 1

Substitute y = 3x - 7 to x^2*y:
x^2 * (3x - 7)
3x^3 - 7x^2
Then differentiate with respect to x, and equate to zero (since we're looking for maximum/minimum):
9x^2 - 14x = 0
x(9x - 14) = 0
x = 0
x = 14/9

Note that if x = 0, the minimum product of x^2 * y is zero.
But if x = 14/9, substituting this back to get y,
y = 3(14/9) - 7
y = 14/3 - 7
y = -7/3
Thus,
x^2 * y
= (14/9)^2 * (-7/3)
= -5.6461 (which is smaller than zero)

hope this helps~ `u`