find the minimum value of x^2*y when x+y=7
Answers
Answer:
Given x+y=2, so Y= 2-x, substituing (2-x) for y in x²+y² gives the expression:
x² + (2-x)² = x² + (4 - 4x +x²) = 2x² -4x+4 which is a parabola
also considered is the fact that a square of a real value is always positive, so the expression x²+y² can be only positive, meaning that the parabola is above X-axis totally.
We need to find the point on the parabola closest to X-axis which will then give the corresponding y value.
parabola equation was y= 2x²–4x+4
x-axis equation is Y=0 (slope =0)
To find closest point on parabola to x-axis, we first find the tangential line on parabola parallel to x-axis i.e parallel to Y=0,
Parabola Y= 2x²–4x+4, so any point on it is of the form (x, 2x²–4x+4), derivative of y: y’ = 2*2x -4 = 4x-4, slope is 0 to be parallel to line y=0, thus 4x-4= 0, resulting in x=1
so x=1 and y= 2–x = 2–1 = 1, (1,1) is the point on parabola closest to x-axis!
x²+y² = 1²+1² = 2 is the minimum value
The minimum value of x²y is 0.
We have to find the minimum value of x²y when x + y = 7.
Here, x + y = 7
⇒y = 7 - x ,
Putting the value of y in x²y , we get
f(x) = x²(7 - x)
⇒f(x) = 7x² - x³
Now differentiating f(x) with respect to x , we get
⇒f'(x) = d(7x² - x³)/dx
= 14x - 3x²
at f'(x) = 0,
⇒ 14x - 3x² = 0
⇒x(14 - 3x) = 0
⇒x = 0, 14/3
Again, differentiating with respect to x, we get
f"(x) = d(14x - 3x²)/dx
⇒f"(x) = 14 - 6x
at x = 0, f"(x) > 0
∴ f(x) will get minima at x = 0
And at x = 14/3 , f"(x) = 14 - 28 < 0 ,
∴ f(x) will get maxima at x = 14/3
now the minimum value of f(x) = x²(7 - x) will be at x = 0.
i.e., f(0) = 0²(7 - 0) = 0