document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1192848>Question 1192848</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi, I am having a lot of trouble with this question. <BR>\n" );
document.write( "Let f(x, y) = x*y2 be a function with the definition set de (x, y) so that<BR>\n" );
document.write( "2x^2+y^2=6<BR>\n" );
document.write( "a) Explain how we can know that f has the definition set of maximum point and minimum point. Explain why f does not have stationary points.<BR>\n" );
document.write( "b) Determine the largest and smallest value of f using the Lagranges method.<BR>\n" );
document.write( "c) What is the answer to parts a) and b) if the definition set is 2x^2+y^2≤6 ?<BR>\n" );
document.write( "d) Let h: R^2 → R be a function given by h(x,y)=4-x^2-y^2<BR>\n" );
document.write( "   i) Find the level curve of f at point P(1,−1). Determine ∇h(1,−1).<BR>\n" );
document.write( "   Draw the point P, as well as the level curve and the gradient to h at this <BR>\n" );
document.write( "   point.<BR>\n" );
document.write( "   ii) Calculate how fast h changes at point P in the direction<BR>\n" );
document.write( "            →   →  →<BR>\n" );
document.write( "            v=−3i+4j<BR>\n" );
document.write( "   iii) In which direction from point P is this change greatest? (Hint:<BR>\n" );
document.write( "   the direction can be given as a vector). </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #848657 by </B><A HREF=/tutors/aboutme.mpl?userid=ElectricPavlov><B>ElectricPavlov(122)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ElectricPavlov><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=848657\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> **a) Existence of Maximum and Minimum Points**\r<BR>\n" );
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document.write( "* **Closed and Bounded Set:** The constraint 2x² + y² = 6 defines an ellipse, which is a closed and bounded set in the plane. <BR>\n" );
document.write( "* **Continuous Function:** The function f(x, y) = x*y² is a polynomial, hence continuous everywhere, and therefore continuous on the ellipse.\r<BR>\n" );
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document.write( "**Extreme Value Theorem:** Since f is continuous on a closed and bounded set (the ellipse), the Extreme Value Theorem guarantees that f attains both a maximum and a minimum value on that set.\r<BR>\n" );
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document.write( "* **No Stationary Points:** <BR>\n" );
document.write( "    * ∇f(x, y) = (y², 2xy) <BR>\n" );
document.write( "    * ∇f(x, y) = 0 only at (0, 0) <BR>\n" );
document.write( "    * The point (0, 0) does not satisfy the constraint 2x² + y² = 6. \r<BR>\n" );
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document.write( "**b) Lagrange Multipliers**\r<BR>\n" );
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document.write( "* **Lagrangian:** <BR>\n" );
document.write( "   L(x, y, λ) = x*y² - λ(2x² + y² - 6) \r<BR>\n" );
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document.write( "* **Partial Derivatives:**<BR>\n" );
document.write( "    * ∂L/∂x = y² - 4λx <BR>\n" );
document.write( "    * ∂L/∂y = 2xy - 2λy <BR>\n" );
document.write( "    * ∂L/∂λ = -(2x² + y² - 6)\r<BR>\n" );
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document.write( "* **System of Equations:**<BR>\n" );
document.write( "    * y² - 4λx = 0 <BR>\n" );
document.write( "    * 2xy - 2λy = 0 <BR>\n" );
document.write( "    * 2x² + y² = 6\r<BR>\n" );
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document.write( "* **Solving the System:**<BR>\n" );
document.write( "    * From the second equation: <BR>\n" );
document.write( "        * 2y(x - λ) = 0 <BR>\n" );
document.write( "        * y = 0 or x = λ <BR>\n" );
document.write( "    * If y = 0, then from the constraint: <BR>\n" );
document.write( "        * 2x² = 6 <BR>\n" );
document.write( "        * x = ±√3 <BR>\n" );
document.write( "        * This gives us the points (±√3, 0) <BR>\n" );
document.write( "    * If x = λ, then from the first equation: <BR>\n" );
document.write( "        * y² - 4x² = 0 <BR>\n" );
document.write( "        * y² = 4x² <BR>\n" );
document.write( "        * Substitute into the constraint: <BR>\n" );
document.write( "            * 2x² + 4x² = 6 <BR>\n" );
document.write( "            * x = ±1 <BR>\n" );
document.write( "        * If x = 1, then y² = 4, so y = ±2 <BR>\n" );
document.write( "        * If x = -1, then y² = 4, so y = ±2 <BR>\n" );
document.write( "        * This gives us the points (1, 2), (1, -2), (-1, 2), and (-1, -2)\r<BR>\n" );
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document.write( "* **Evaluate f at the Critical Points:**<BR>\n" );
document.write( "    * f(√3, 0) = 0 <BR>\n" );
document.write( "    * f(-√3, 0) = 0 <BR>\n" );
document.write( "    * f(1, 2) = 4 <BR>\n" );
document.write( "    * f(1, -2) = 4 <BR>\n" );
document.write( "    * f(-1, 2) = -4 <BR>\n" );
document.write( "    * f(-1, -2) = -4\r<BR>\n" );
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document.write( "* **Conclusion:** <BR>\n" );
document.write( "    * **Maximum Value:** 4 at (1, 2) and (1, -2) <BR>\n" );
document.write( "    * **Minimum Value:** -4 at (-1, 2) and (-1, -2)\r<BR>\n" );
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document.write( "**c) Definition Set: 2x² + y² ≤ 6**\r<BR>\n" );
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document.write( "* **Interior Points:** We already determined that there are no stationary points within the interior of the ellipse.<BR>\n" );
document.write( "* **Boundary:** The analysis in part (b) already considered the boundary (2x² + y² = 6).\r<BR>\n" );
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document.write( "* **Conclusion:** The maximum and minimum values remain the same as in part (b) because the boundary points still provide the extrema.\r<BR>\n" );
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document.write( "**d) Function h(x, y) = 4 - x² - y²**\r<BR>\n" );
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document.write( "**i) Level Curve and Gradient at P(1, -1)**\r<BR>\n" );
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document.write( "* **Level Curve:** <BR>\n" );
document.write( "    * f(1, -1) = 1 <BR>\n" );
document.write( "    * The level curve of f at P(1, -1) is the set of points (x, y) such that f(x, y) = 1, which is the curve x*y² = 1.\r<BR>\n" );
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document.write( "* **Gradient of h:** <BR>\n" );
document.write( "    * ∇h(x, y) = (-2x, -2y) <BR>\n" );
document.write( "    * ∇h(1, -1) = (-2, 2)\r<BR>\n" );
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document.write( "**ii) Directional Derivative**\r<BR>\n" );
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document.write( "* **Unit Vector in the Direction of v:** <BR>\n" );
document.write( "    * ||v|| = √((-3)² + 4²) = 5 <BR>\n" );
document.write( "    * <BR>\n" );
document.write( " </SPAN>\n" );
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