If \(\displaystyle y\) varies directly with the square root of \(\displaystyle x\), then for some constant of variation \(\displaystyle K\), 

\(\displaystyle \small y = K \sqrt{x}\)

If \(\displaystyle x = 25\), then \(\displaystyle y = 48\); therefore, the equation becomes 

\(\displaystyle \small \small 48 = K \sqrt{25}\), 

or

\(\displaystyle 5K = 48\).

Divide by 5 to get \(\displaystyle K = 9.6\), making the equation 

\(\displaystyle \small \small y = 9.6 \sqrt{x}\).

If \(\displaystyle x = 81\), then \(\displaystyle \small \small \small y = 9.6 \sqrt{81} = 9.6 \cdot 9 = 86.4\).

Since \(\displaystyle y\) varies directly with \(\displaystyle x^{3}\), \(\displaystyle y=Kx^{3}\) where \(\displaystyle K\) is a constant.

1. Solve for \(\displaystyle K\) when \(\displaystyle x=5\) and \(\displaystyle y=61.125\).

\(\displaystyle y=Kx^{3}\)

\(\displaystyle 61.125=K(5^{3})\)

\(\displaystyle K=\frac{61.125}{125}=0.489\)

2. Use your equation to solve for \(\displaystyle y\) when \(\displaystyle x=11\).

\(\displaystyle y=0.489x^{3}\)

\(\displaystyle y=0.489(11^{3})=650.859\)

Since \(\displaystyle y\) varies indirectly with \(\displaystyle \sqrt{x}\), \(\displaystyle y=\frac{K}{\sqrt{x}}\)

1. Solve for \(\displaystyle K\) when \(\displaystyle y=12.3\) and \(\displaystyle x=4\).

\(\displaystyle y=\frac{K}{\sqrt{x}}\)

\(\displaystyle 12.3=\frac{K}{\sqrt{4}}\)

\(\displaystyle 12.3= \frac{K}{2}\)

\(\displaystyle K=24.6\)

2. Use the equation you found to solve for \(\displaystyle y\) when \(\displaystyle x=17\).

\(\displaystyle y=\frac{24.6}{\sqrt{x}}\)

\(\displaystyle y=\frac{24.6}{\sqrt{17}}=5.97\)

1. Since \(\displaystyle y\) varies directly with \(\displaystyle x^{2}\): 

\(\displaystyle y=Kx^{2}\) with K being some constant.

2. Solve for K using the x and y values given:

\(\displaystyle 75=K(5^{2})\)

\(\displaystyle 75=K(25)\)

\(\displaystyle K=3\)

3. Use the equation you solved for to find the value of y:

\(\displaystyle y=3x^{2}\)

\(\displaystyle y=3(10^{2})\)

\(\displaystyle y=300\)

1. Since \(\displaystyle y\) varies indirectly with \(\displaystyle x^{3}\):

\(\displaystyle y=\frac{K}{x^{3}}\)

2. Use the given x and y values to determine the value of K:

\(\displaystyle 2=\frac{K}{2^{3}}\)

\(\displaystyle 2=\frac{K}{8}\)

\(\displaystyle K=16\)

3. Using the equation along with the value of K, find the value of y when x=5:

\(\displaystyle y=\frac{16}{x^{3}}\)

\(\displaystyle y=\frac{16}{5^{3}}\)

\(\displaystyle y=0.128\)

1. Since \(\displaystyle y\) varies directly with \(\displaystyle \sqrt[3]{x}\):

\(\displaystyle y=K\sqrt[3]{x}\)

2. Use the values given for x and y to solve for K:

\(\displaystyle y=K\sqrt[3]{x}\)

\(\displaystyle 3=K\sqrt[3]{8}\)

\(\displaystyle 3=K(2)\)

\(\displaystyle K=\frac{3}{2}\)

3. Use your new equation with the K you solved for to solve for y when x=27:

\(\displaystyle y=(\frac{3}{2})\sqrt[3]{27}\)

\(\displaystyle y=(\frac{3}{2})(3)=4.5\)

1. Since y varies indirectly with \(\displaystyle \sqrt{x}\):

\(\displaystyle y=\frac{K}{\sqrt{x}}\)

2. Solve for K using the x and y values given:

\(\displaystyle 0.5=\frac{K}{\sqrt{4}}\)

\(\displaystyle 0.5=\frac{K}{2}\)

\(\displaystyle K=1\)

3. Using the equation you created by solving for K, find y when x=100:

\(\displaystyle y=\frac{1}{\sqrt{x}}\)

\(\displaystyle y=\frac{1}{\sqrt{100}}\)

\(\displaystyle y=\frac{1}{10}\)

Using our two known points, we can use interpolation to determine the value at any point between them with the following formula:

\(\displaystyle \frac{y_2-y_1}{x_2-x_1}=\frac{y-y_1}{x-x_1}\)

Where \(\displaystyle (x_1,y_1)\) is our first given point, \(\displaystyle (x_2,y_2)\) is our second given point, and \(\displaystyle (x,y)\) is the point we want to find. We know our two given points, as well as the x value of our unknown point, so now all we must do is plug in all of our known values and solve for y, our only unknown:

\(\displaystyle \frac{93-81}{25-24}=\frac{y-81}{24.75-24}\)

\(\displaystyle 12=\frac{y-81}{0.75}\)

\(\displaystyle y-81=9\rightarrow y=f(24.75)=90\)

We want to do a linear interpolation since the relationship between workers and units can be assumed to be linear. This means there is a constant slope between the points, so the slope between two known points will be equal to the slope between the point we are trying to find and some known point. This is expressed in the relation:

\(\displaystyle \frac{y-y_0}{x-x_0}=\frac{y_1-y_0}{x_1-x_0}\),

where \(\displaystyle y\) and \(\displaystyle x\) are the points we want to find and \(\displaystyle (x_1,y_1)\) and \(\displaystyle (x_0,y_0)\) are known. We choose the known points to be those that are just to the left and right of the point we are trying to find,

\(\displaystyle (100,1300)\) and \(\displaystyle (200,2300)\).

Plugging these into our interpolation formula and knowing \(\displaystyle x=110\), we can find \(\displaystyle y\), the units output per day.

\(\displaystyle \frac{y-1300}{110-100}=\frac{2300-1300}{200-100}\).

Simplifying and rearranging to solve for \(\displaystyle y\):

\(\displaystyle y=1400\).

So there are \(\displaystyle 1400\) units produced when the number of workers is \(\displaystyle 110\).

Use the formula for interpolation to determine the value of y:

\(\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{x-x_{1}}\)

We will use (30, 51) as our x2 and y2 and (20, 36) as our x1 and y1 and we will solve for y using 26.5 for x.

\(\displaystyle \frac{51-36}{30-20}=\frac{y-36}{26.5-20}\)

\(\displaystyle \frac{15}{10}=\frac{y-36}{6.5}\)

\(\displaystyle \frac{3}{2}(6.5)=y-36\)

\(\displaystyle y=45.75\)