For what value of k will the points of simultaneous equations have no solution?
1 answer:
You might be interested in
Answer:
92
Step-by-step explanation:
Since the sum of the two numbers is 5, we can represent one of them by x and the other by 5-x. Then the desired product is ...
x²(5-x)³
A graphing calculator can show the extreme values of this on the interval 1 ≤ x ≤ 4. The maximum is 108 at x=2; the minimum is 16 at x=4.
The difference between the maximum and minimum is 108-16 = 92.
_____
If you like, you can take the derivative and set it to zero.
f(x) = x²(5 -x)³
f'(x) = 2x(5 -x)³ +x²(-3)(5-x)² = x(5 -x)²(2(5-x) -3x)
f'(x) = 5x(5-x)²(2-x)
This will be zero for x=0, x=5, and x=2. The points at x=0 and x=5 represent minima in the product. The values x=0 and x=5 are not in the domain of interest. The point at x=2 represents a maximum.
To find the function extremes on an interval, we need to evaluate the function where the derivative is zero, and also at the ends of the interval. So, the function values of interest are ...
f(1) = 1²·4³ = 64
f(2) = 2²·3³ = 108 . . . . product maximum
f(4) = 4²·1³ = 16 . . . . . . product minimum
The difference between the maximum and minimum is 92.
