Let z=2x+5y−−−−−−√
. Then:
The rate of change in z
at (2,4)
as we change x
but hold y
fixed is
, and
The rate of change in z
at (2,4)
as we change y
but hold x
fixed is
To find the rate of change in z at (2,4) as we change x but hold y fixed, we take the partial derivative of z with respect to x and evaluate it at (2,4).
The partial derivative of z with respect to x is given by:
∂z/∂x = 2
So, the rate of change in z at (2,4) as we change x but hold y fixed is 2.
To find the rate of change in z at (2,4) as we change y but hold x fixed, we take the partial derivative of z with respect to y and evaluate it at (2,4).
The partial derivative of z with respect to y is given by:
∂z/∂y = 5/2√(2x+5y)
Plugging in the values of x=2 and y=4 into the partial derivative, we get:
∂z/∂y = 5/2√(2(2)+5(4)) = 5/2√(4+20) = 5/2√24 = 5/2√6
So, the rate of change in z at (2,4) as we change y but hold x fixed is 5/2√6.
To find the rate of change in z at (2,4) as we change x but hold y fixed, we need to evaluate ∂z/∂x at x = 2 and y = 4.
Taking the partial derivative of z with respect to x, we get:
∂z/∂x = 2√(2x + 5y)
Substituting x = 2 and y = 4, we have:
∂z/∂x = 2√(2(2) + 5(4))
= 2√(4 + 20)
= 2√24
= 2(2√6)
= 4√6
Therefore, the rate of change in z at (2,4) as we change x but hold y fixed is 4√6.
To find the rate of change in z at (2,4) as we change y but hold x fixed, we need to evaluate ∂z/∂y at x = 2 and y = 4.
Taking the partial derivative of z with respect to y, we get:
∂z/∂y = 5√(2x + 5y)
Substituting x = 2 and y = 4, we have:
∂z/∂y = 5√(2(2) + 5(4))
= 5√(4 + 20)
= 5√24
= 5(2√6)
= 10√6
Therefore, the rate of change in z at (2,4) as we change y but hold x fixed is 10√6.