Consider the following. (Round your answers to four decimal places.)f(x, y) = yex(a) Evaluate f(2, 1) and f(2.5, 1.65) and calculate Δz(b) Use the total differential dz to approximate Δz.

Step-by-step explanation:

Given the function f(x,y) = ye^x

To evaluate f(2,1), we will substitute x = 2 and y = 1 into the given function to have;

f(2,1) = 1×e^2

f(2,1) = e^2

f(2,1) = 7.3891(to 4dp)

For f(2.5, 1.65), x = 2.5 and y =1.65

f(2.5, 1.65) = 1.65e^2.5

f(2.5, 1.65) = 1.65×12.1825

f(2.5, 1.65) = 15.2281 (to 4dp).

∆z = f(2.5, 1.65) - f(2, 1)

∆z = 15.2281-7.3891

∆z = 7.8390

dz = fxdx + fydy

fx is the differential of the function with respect to x keeping y constant.

Since f(x,y) = ye^x

fx = ye^x + 0e^x

fx = ye^x

dx = x2-x1 = 2.5-2

dx = 0.5

fy = e^x

dy = (y2-y1) = 1.65-1

dy = 0.65

Using the point (2, 1) for x and y and substituting the values gotten into dz function:

dz =ye^x(0.5) + e^x(0.65)

If x = 2 and y = 1

dz = 1e^2(0.5) + e^2(0.65)

dz = 7.3891(0.5)+7.3891(0.65)

dz = 7.3891(0.5+0.65)

dz = 7.3891(1.15)

dz = 8.4975

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