Question

The height of a certain hill (in feet) is given by

$\mathit{h}\left(x,y\right)\mathbf{=}\mathbf{10}\left(2xy-3x^2-4y^2-18x+28y+12\right)$

Where y is the distance (in miles) north, x the distance east of South Hadley.

(a) Where is the top of hill located?

(b) How high is the hill?

(c) How steep is the slope (in feet per mile) at a point 1 mile north and one mileeast of South Hadley? In what direction is the slope steepest, at that point?

Short answer

(a) The hill is 3 miles west and 2 miles to north.

(b) Theheight of the hill is 720 ft.

(c) The slope is $311\frac{\text{ft}}{\text{mile}}$, at the point (1, 1) and in the north west direction.

Step-by-step solution

Write the given information. 

Write the given function.

$h\left(x,y\right)=10\left(2xy-3x^2-4y^2-18x+28y+12\right)$

Define gradient of the function.

The gradient of the function is defined as its slope on the curve of that function for the given particular point.

Consider the function.

$f\left(x,y,z\right)$

Then, the gradient of the function is,

$\nabla \mathbf{f}=\frac{\partial \mathbf{f}}{\partial \mathbf{x}}\hat{x}\mathbf{+}\frac{\partial \mathbf{f}}{\partial \mathbf{z}}\hat{y}\mathbf{+}\frac{\partial \mathbf{f}}{\partial \mathbf{z}}\hat{z}$

Determine the coordinates for the top of the hill.

(a)

Write the given function.

$h\left(x,y\right)=10\left(2xy-3x^2-4y^2-18x+28y+12\right)$ ……. (1)

Differentiate the function with respect to x and equate it equal to 0.

$$\begin{gathered} \frac{dh}{dx}=0 \\ 20y-60x-180=0 \end{gathered}$$

Differentiate the function with respect to y and equate it equal to 0.

$$\begin{gathered} \frac{dh}{dy}=0 \\ 20x-80y+280=0 \end{gathered}$$

From the two equations.

$$\begin{gathered} x=-2 \\ y=3 \end{gathered}$$

Therefore, the hill is 3 miles west and 2 miles to north.

Determine the height of the hill.

(b)

Substitute for x and 3 for y in the equation (1).

$\begin{array}{rcl}h\left(x,y\right)& =& 10\left(2\left(-2\right)\left(3\right)-3{\left(-2\right)}^2-4{\left(3\right)}^2-18\left(-2\right)+28\left(3\right)+12\right)\\ & =& 720\text{ft}\\ & & \end{array}$

Therefore, the height of the hill is 720 ft.

Determine the steepness, direction and point of the slope.

(c)

Determine the gradient of the equation (1).

$\begin{array}{rcl}\nabla h\left(x,y\right)& =& \frac{\partial h}{\partial x}\hat{x}+\frac{\partial h}{\partial y}\hat{y}\\ & =& \left(20y-60x-180=0\right)\hat{x}+\left(20x-80y+280\right)\hat{y}\\ & =& {\left[{\left\{20\left(y-3x+9\right)\right\}}^2+20{\left(x-4y+14\right)}^2\right]}^{\frac{1}{2}}\\ & & \end{array}$

Substitute1 for x and 1 for yin the equation.

$\begin{array}{rcl}\nabla h\left(x,y\right)& =& {\left[{\left\{20\left(\left(1\right)-3\left(1\right)+9\right)\right\}}^2+20{\left(\left(1\right)-4\left(1\right)+14\right)}^2\right]}^{\frac{1}{2}}\\ & =& 311\frac{\text{ft}}{\text{mile}}\\ & & \end{array}$

Therefore, the slope is$311\frac{\text{ft}}{\text{mile}}$ , at the point (1, 1) and in the north west direction.