91Ó°ÊÓ

Write the given function.

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

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,

$∇\mathbf{f}=\frac{∂\mathbf{f}}{∂\mathbf{x}}\hat{x}\mathbf{+}\frac{∂\mathbf{f}}{∂\mathbf{z}}\hat{y}\mathbf{+}\frac{∂\mathbf{f}}{∂\mathbf{z}}\hat{z}$

(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.

(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.

(c)

Determine the gradient of the equation (1).

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

Substitute1 for x and 1 for yin the equation.

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

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