Consider a bivariate function f(x,y) such that f(1,1)=2000, f_x(1,1)=4.2 and f_y(1,1)=5.0. Estimate the value of f(2,3) by the partial derivatives. Round your answer to 2 decimal places.

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We start by identifying the given information and how to apply a linear approximation around the point (1,1).

- The function value at (1,1) is f(1,1) = 2000.
- The partial derivatives at (1,1) are f_x(1,1) = 4.2 and f_y(1,1) = 5.0.
- We want to estimate f at (x,y) = (2,3). The changes in coordinates are Δx = 2 - 1 = 1 and Δy = 3 - 1 = 2.

Using the first-order (linear) approximation for a multivariable function:
f(x,y) ≈ f(1,1) + f_x(1,1)·Δx + f_y(1,1)·Δy.

Substitute the known values:
f(2,3) ≈ 2000 + (4.2)(1) + (5.0)(2) = 2000 + 4.2 + 10.

Compute the sum:
2000 + 4.2 + 10 = 2014.2.

Round to two decimals as requested: 2014.20.

To check for consistency, note that the estimate uses only the linear (tangent plane) approximation, which is most accurate for points close to (1,1). Since Δx and Δy are modest (1 and 2), the result follows directly from the given derivative information. The result should be interpreted as an approximation, not an exact value, unless higher-order terms are included.

Consider a bivariate function f(x,y) such that f(1,1)=2000, f_x(1,1)=4.2 and f_y(1,1)=5.0. Estimate the value of f(2,3) by the partial derivatives. Round your answer to 2 decimal places.

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