Question 14.7.1: Consider Example 14.6.4, and suppose we change the first con......
Consider Example 14.6.4, and suppose we change the first constraint to x + 2y +z = 31 and the second constraint to 2x − y − 3z = 9. Estimate the corresponding change in the value function by using (14.7.4). Find also the new exact value of the value function.
f^{*}(\mathbf{c}+\operatorname*{d}\!\mathbf{c})-f^{*}(\mathbf{c})\approx\lambda_{1}(\mathbf{c})\,\mathrm{d}c_{1}+\mathbf{\cdot\cdot\cdot}+\lambda_{m}(\mathbf{c})\,\mathrm{d}c_{m} (14.7.4)
Using the notation introduced above and the results in Example 14.6.4, we have c_{1} = 30, c_{2} = 10, dc_{1} = 1, dc_{2} = −1, λ_{1}(30, 10) = 12, λ_{2}(30, 10) = 4, and f^{∗}(c_{1}, c_{2}) = f^{∗}(30, 10) = 10^{2} + 10^{2} + 0^{2} = 200. Then, approximation (14.7.4) yields
f^{*}(30+1,10-1)-f^{*}(30,10)\approx\lambda_{1}(30,10)\,\mathrm{d}c_{1}+\lambda_{2}(30,10)\,\mathrm{d}c_{2}\quad\quad\quad=12\cdot1+4\cdot(-1)\ =\ 8
Thus, f^{*}(31,9)\approx200+8=208.
To find the exact value of f^{∗}(31, 9), observe that (vi) in Example 14.6.4 is still valid. Thus, we have the three equations x + 2y + z = 31, 2x − y − 3z = 9, x − y + z = 0, whose solutions for x, y, and z are 151/15, 31/3, and 4/15, respectively. We find that f^{∗}(31, 9) =15 614/75 ≈ 208.19.