Which point of the mountain

Question

A mountaineer wishes to make a descent from a mountain, starting at the top of the mountain at $(0, 0)$. The height of the mountain at position (x, y) is given by $$h(x, y) = 3000 - \frac{1}{10000}(5x^2+4xy+2y^2)$$ In 30 minutes, the mountaineer can arrive at any point that lies on a circle with a radius 1000 and centered at $(0, 0)$. Which point should the mountainer target if he is to descent as much as possible?

Using Lagrange function

Answer 1

You can simplify the algebra a bit by realizing that you are really just looking to maximize $$f(x,y) = 5x^2+2xy+2y^2 $$ subject to the constraint $$g(x,y)=x^2+y^2=1000$$ so that $$f(x,y) = 3x^2+2xy+2000 $$

Answer 2

HINT

We need to find the minimum of $h(x,y)$ with the constraint $x^2+y^2=1000$.

To solve we need to apply Lagrange multipliers for $x^2+y^2=1000$, that is solve the system

• $h_x=2\lambda x$
• $h_y=2\lambda y$
• $x^2+y^2=1000$

Answer 3

Here we got , $$3000-\frac{2 {{y}^{2}}+4 x y+5 {{x}^{2}}}{10000}=\lambda\, \left( {{y}^{2}}+{{x}^{2}}-1000000\right)........(1) $$ $$-\frac{y}{2500}-\frac{x}{1000}=2 \lambda x ........(2)$$ $$-\frac{y}{2500}-\frac{x}{2500}=2 \lambda y ........(3)$$ solving for the minimum value,we get- $$\lambda=-\frac{3}{5000}$$ $$x=\pm400\sqrt{5}$$ $$y=\pm200\sqrt{5}$$ Hence, $$h=2400 m $$