Find the value of constant(a+b+c)so that the directional derivative of the functionf=axy2+byz+cz2x3at the point (1, 2, –1) has maximum magnitude 64 in the direction parallel toyaxis :Correct answer is '-20'. Can you explain this answer?

Question

Find the value of constant(a+b+c)so that the directional derivative of the functionf=axy2+byz+cz2x3at the point (1, 2, &ndash;1) has maximum magnitude 64 in the direction parallel toyaxis :Correct answer is '-20'. Can you explain this answer?

Answer

To find the value of the constant (a, b, c) that would give the desired directional derivative, we need to use the formula for the directional derivative:

D_v(f) = ∇f · v

where D_v(f) is the directional derivative of the function f, ∇f is the gradient of f, and v is the direction vector.

First, let's find the gradient of f. The gradient of f is given by:

∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z)

Taking the partial derivatives of f with respect to x, y, and z, we get:

∂f/∂x = ay^2 z c z^2
∂f/∂y = ax z^2 c z^2
∂f/∂z = axy^2 cz^2

Now, let's find the direction vector v. The direction vector is given by the components (a, b, c).

Using the formula for the directional derivative, we have:

D_v(f) = ∇f · v
= (∂f/∂x, ∂f/∂y, ∂f/∂z) · (a, b, c)
= (ay^2 z c z^2, ax z^2 c z^2, axy^2 cz^2) · (a, b, c)
= a(ay^2 z c z^2) + b(ax z^2 c z^2) + c(axy^2 cz^2)
= a^2 y^2 z^3 c + abx z^4 c + acx y^2 z^4 c^2

To find the value of the constant (a, b, c), we need to set the directional derivative equal to the desired value. Let's say the desired directional derivative is D_v(f) = k.

k = a^2 y^2 z^3 c + abx z^4 c + acx y^2 z^4 c^2

Since we are given the point (1, 2, 3), let's substitute these values into the equation:

k = a^2 (2)^2 (3)^3 c + a(1)(3)^4 c + a(1)(2)^2 (3)^4 c^2
k = 36a^2 c + 81ac + 5832ac^2

Now, we can solve this equation for the constant (a, b, c) by setting it equal to the desired value of k and solving for (a, b, c).

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