Question

Find all values of p such that 2(x+4)(x-2p) has a minimum value of -18? I know x=-4 but I do not know that helps us. Please explain in steps!

The temperature of a point (x,y) in the plane is given by the expression x^2-y^2-4x+2y. What is the temperature of the coldest point on the plane? I know we can factor this to x(x-4)-y(y-2) but I am not sure what to do from there.

Thank you!

Answer 1

you know the minimum value is midway between the roots, which are -4 and 2p.

So, the vertex is at x = (-4+2p)/2 = p-2

So, now we want the y-coordinate of the vertex, which is

y = 2(p-2+4)(p-2-2p) = 2(p+2)(-p-2)
= -2(p+2)^2

If that is -18, then
-2(p+2)^2 = -18
p+2 = ±3, so
p = -1 or 5

You can plug those is and verify that the vertex is at y = -18

Note that

f(x,y) = (x^2-4x) + (y^2+2y)
= (x^2-4x+4) + (y^2+2y+1) - (4+1)
= (x-2)^2 + (y+1)^2 - 5

Clearly the minimum is achieved at (2,-1) where f(2,-1) = -5

For all other values of x and y, the first terms are positive.