b2 - 5.18 (optimering)

Svaret enligt facit ska vara F_max=1, F_min=-2.

Har gått igenom min beräkningar men hittar inte vart jag gått fel?$\begin{array}{l}\text{Lös: }f\left(x, y\right)=x{y}^2-x^2-y, 0\le x\le 1, 0\le y\le 2. \\ \text{1. Stationära punkter:}\\ f^{\text{'}}x=y^2-2x=0, \left[x=\frac{1}{2}{y}^2\right]\\ f^{\text{'}}y=2xy-1=0.\\ 2\left(\frac{1}{2}{y}^2\right)y-1=0, \left[y=1\right], \left[x=\frac{1}{2}\right].\\ \text{Stationär punkt:} \left(x, y\right)=\left(\frac{1}{2}, 1\right),\\ \text{Intressant punkt:} f\left(\frac{1}{2}, 1\right)=\left[-\frac{3}{4}\right].\\ \text{2. Randen:}\\ g_1\left(x\right)=f\left(x, 0\right)=\left[-x^2\right], g_1\left(x\right)=-x^2=0, \left[x=0\right].\\ \text{Intressant punkt:} f\left(0, 0\right)=\left[0\right].\\ g_2\left(x\right)=f\left(x, 2\right)=4x-x^2-2, g_2\left(x\right)=x^2-4x+2=0\\ \text{PQ−formel:} x=2+-{\left(4-2\right)}^{\frac{1}{2}}=2+-2^{\frac{1}{2}}, \left[x_1=2+2^{\frac{1}{2}}\right]\left[x_2=2-2^{\frac{1}{2}}\right]\\ \text{Intressant punkt:} f\left(2+2^{\frac{1}{2}}, 2\right)=\left(2+2^{\frac{1}{2}}\right){\left(2\right)}^2-{\left(2+2^{\frac{1}{2}}\right)}^2-2=0 \left(2+2^{\frac{1}{2}}>1, ej intressant punkt\right)\\ \text{Intressant punkt:} f\left(2-2^{\frac{1}{2}}, 2\right)=\left(2-2^{\frac{1}{2}}\right){\left(2\right)}^2-{\left(2-2^{\frac{1}{2}}\right)}^2=\left[2\right]\\ g_3\left(x\right)=f\left(0, y\right)=-y, g_3\left(x\right)=-y=0, \left[y=0\right]. \left(f\left(0, 0\right) redan uträknat\right)\\ g_4\left(x\right)=f\left(1, y\right)=y^2-1-y, g_4\left(x\right)=y^2-y-1=0, \\ \text{PQ−formel:} y=\frac{1}{2}+-{\left(\frac{1}{4}+1\right)}^{\frac{1}{2}}=\frac{1}{2}+-\frac{3}{2}, \left[y_1=2\right], \left[y_2=-1\right].\\ \text{Intressant punkt:} f\left(1, 2\right)=\left[1\right],\\ \text{Intressant punkt:} f\left(1, -1\right) \left(-1<0, ej intressant punkt\right).\\ \text{3. Hörn: }\\ f\left(0, 0\right) \left(redan uträknat\right)\\ f\left(0, 2\right)=\left[-2\right]\\ f\left(1, 2\right) \left(redan uträknat\right)\\ f\left(1, 0\right)=\left[-1\right]\\ \text{Alla intressanta punkter:}\\ f\left(\frac{1}{2}, 1\right)=\left[-\frac{3}{4}\right], f\left(0, 0\right)=\left[0\right],\\ f\left(2-2^{\frac{1}{2}}, 2\right)=\left[2\right], f\left(1, 2\right)=\left[1\right],\\ f\left(0, 2\right)=\left[-2\right], f\left(1, 0\right)=\left[-1\right].\\ \text{Svar: }{f}_{\min }=-2, f_{\max }=2.\end{array}$