Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ellipse StartFraction x squared Over 25 EndFraction plusStartFraction y squared Over 16 EndFraction equals1 with sides parallel to the coordinate axes.

Answer with Step-by-step explanation: Let  a rectangle box whose dimensions are u and v.Then, [tex](u/2, v/2)[/tex]must lie on the ellipseGiven equation of ellipse [tex]\frac{x^2}{25}+\frac{y^2}{16}=1[/tex][tex] (u/2,v/2)[/tex]must lie on the circle therefore, [tex]\frac{u^2}{100}+\frac{v^2}{64}=1[/tex]with [tex]u,v\geq 0[/tex]Suppose , we have  to maximize a  function [tex]f(u,v)=uv[/tex] subject to constraints g(u,v)=[tex]\frac{u^2}{100}+\frac{v^2}{64}=1[/tex][tex]\nabla f=<v,u> \nabla g=<\frac{u}{50},\frac{v}{32} >[/tex]Using Lagrange multipliers method [tex]\nabla f=\lambda\nabla g[/tex][tex]v=\lambda \frac{u}{50}[/tex][tex]u=\lambda\frac{v}{32}[/tex][tex]\lambda=\frac{50v}{u}=\frac{32u}{v}[/tex][tex]50v^2=32u^2[/tex]if u=0 then v=0 g(u,v)=[tex]0\neq 1[/tex]It is absurd condition.Therefore, we take u and v >0[tex]v^2=\frac{32u^2}{50}=\frac{16}{25}u^2[/tex]Substitute the value in ellipse equation then we get [tex]\frac{u^2}{100}+\frac{u^2}{100}=1[/tex][tex]\frac{2u^2}{100}=1[/tex][tex]u^2=50[/tex][tex]u=5 \sqrt2[/tex][tex]v^2=\frac{16}{25}\times 50=32[/tex][tex]v=4\sqrt2[/tex]The critical point is ([tex]5\sqrt2, 4\sqrt2)[/tex].Therefore, we concluded that the dimensions of the rectangle of greatest area is attained by choosing a box of dimensions [tex]5\sqrt2 \times 4\sqrt2[/tex].