In order to Problem Solve with Quadratics more easily, we should first identify what we need to solve or look for. These variables generally include:
Identifying the key variable will allow us to determine what form to put the original expression in.
If we wanted to identify the max or min points, we should convert the expression to Vertex Form since its \(k\) value tells us the max/min. Likewise, if we wanted to find the roots (or \(x\)-intercepts), we can convert the expression to Factored Form since its \(p\) and \(q\) values show the roots. Lastly, if we wanted to find the \(y\)-intercept, we can convert the expression to Standard Form since its \(c\) value represents the \(y\)-intercept.
A harbour ferry service has about \(240 000\; [\text{riders/day}]\) for a fare of \($2\). The port authority wants to increase the fare to help with increasing operational costs. Research has shown that for every \($0.10\) increase in the fare the number of riders will drop by \(10 000\).
i. First, we can determine the respective formulas for determining Price and Quantity/People:
\(P = 2 + 0.10x\)
\(Q = 240 000 - 10 000x\)
Next, we can use the Revenue formula to determine what the Revenue equation will be:
\(\text{Revenue} = \text{Price} \times \text{Quantity}\)
OR
\(R = P \times Q\)
As we have polynomials for Price and Quantity/People, we can expand these using FOIL to determine the final equation:
\(R = (2 + 0.10x)(240 000 - 10 000x)\)
\(R = (2)(240 000) + (2)(-10 000x) + (0.10x)(240 000) + (0.10x)(-10 000x)\)
\(R = 480 000 - 20 000x + 24 000x - 1000x²\)
\(R = -1000x² + 4000x + 480 000\)
Therefore, we can represent Revenue algebraically as \(\boldsymbol{R = -1000x² + 4000x + 480 000}\).
ii. Finding the maximum revenue is similar to finding the vertex for a parabola. In order to do this, we can find the respective roots of the original equations:
We can start by determining the first root:
\(\text{root}_1: 2 + 0.10x\)
\(\text{root}_1: 0.10x = -2\)
\(\text{root}_1: \cfrac{\cancel{0.10}x}{\cancel{0.10}} = \cfrac{-2}{0.10}\)
\(\text{root}_1: x = -20\)
Next, we can determine the second root:
\(\text{root}_2: 240 000 - 10 000x\)
\(\text{root}_2: 10 000x = 240 000\)
\(\text{root}_2: \cfrac{\cancel{10 000}x}{\cancel{10 000}} = \cfrac{240 000}{10 000}\)
\(\text{root}_2: x = 24\)
Now that we have the roots, we can divide the sum in half to determine the \(x\)-coordinate of the vertex. This value represents the # of times the fare should be increased to maximize the revenue:
\(x = \cfrac{-20 + 24}{2}\)
\(x = 2\)
Therefore, we can determine that the fare should get increased \(\boldsymbol{2}\) times to maximize the revenue.
iii. To determine the new fare, we need to plug the \(x\)-coordinate, \(2\), into the Price equation:
\(P = 2 + 0.10x\)
\(P = 2 + (0.10)(2)\)
\(P = 2 + 0.20\)
\(P = 2.20\)
Therefore, we can determine that the new fare that maximizes the revenue is \(\boldsymbol{$2.20}\).
iv. To determine the # of riders for the maximum revenue, we need to plug the \(x\)-coordinate into the Quantity equation:
\(Q = 240 000 - 10 000x\)
\(Q = 240 000 - (10 000)(2)\)
\(Q = 240 000 - 20 000\)
\(Q = 218 000\)
Therefore, we can determine that the # of riders needed for maximum revenue is \(\boldsymbol{218 000}\).
v. We can determine the maximum revenue by plugging the \(x\)-coordinate into the Revenue equation:
\(R = -1000x² + 4000x + 480 000\)
\(R = (-1000)(2)² + (4000)(2) + 480 000\)
\(R = (-1000)(4) + 8000 + 480 000\)
\(R = -4000 + 8000 + 480000\)
\(R = $484 000\)
Therefore, we can determine that the maximum revenue is \(\boldsymbol{$484 000}\).
We can start by identifying the equation of the base, based on the description in the question:
We can now plug this new equation into the Area of a Triangle formula:
\(A = \cfrac{1}{2}bh\)
\(308 = \cfrac{1}{2}(3h + 2)h\)
\(308 = \cfrac{1}{2}(3h² + 2h)\)
We can cross-multiply and shift all terms onto one side in order to get a standard quadratic equation:
\((308)(2) = \left(\cfrac{1}{2}(3h² + 2h)\right)(2)\)
\(616 = 3h² + 2h\)
\(3h² + 2h - 616 = 0\)
We can now use the quadratic formula to determine the height. We can identify that \(\textcolor{red}{a = 3}\), \(\textcolor{green}{b = 2}\) and \(\textcolor{blue}{c = -616}\):
\(x = \cfrac{-\textcolor{green}{b} +- \sqrt{\textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c}}}{2\textcolor{red}{a}}\)
\(x = \cfrac{-(\textcolor{green}{2}) +- \sqrt{(\textcolor{green}{2})^2 - 4(\textcolor{red}{3})(\textcolor{blue}{-616})}}{2(\textcolor{red}{3})}\)
\(x = \cfrac{-2 \pm \sqrt{4 + 7392}}{6}\)
\(x = \cfrac{-2 \pm \sqrt{7396}}{4}\)
Using \(\pm\), we can determine the height either by adding or subtracting the square root value.
We can determine the first potential height by adding the square root value:
\(h_1 = \cfrac{-2 + 86}{6}\)
\(h_1 = \cfrac{84}{6}\)
\(h_1 = 14\;[\text{cm}]\)
We can determing the second potential height by subtracting the square root value:
\(h_2 = \cfrac{-2 - 86}{6}\)
\(h_2 = \cfrac{-88}{6}\)
\(h_2 = -14.67\;[\text{cm}]\)
Since the height can't be negative. we can determine that \(h = 14\;[\text{cm}]\). We can plug this value into the base equation to determine the base value:
\(b = 3h + 2\)
\(b = 3(14) + 2\)
\(b = 42 + 2\)
\(b = 44\;[\text{cm}]\)
Therefore, we can determine that \(\boldsymbol{h = 14 [\textbf{cm}]}\) and \(\boldsymbol{b = 44 [\textbf{cm}]}\).
First, we can represent each of these values as \(x\), \(x + 1\), \(x + 2\) and \(x + 3\) respectively. We can then place these values into a single equation:
We can expand this equation, collect like terms and move all terms onto one side to determine the standard quadratic equation:
\(x² + (x + 1)(x + 1) + (x + 2)(x + 2) + (x + 3)(x + 3) = 630\)
\(x² + x² + 2x + 1 + x² + 4x + 4 + x² + 6x + 9 = 630\)
\(4x² + 12x + 14 = 630\)
\(4x² + 12x -616 = 0\)
We can now use the quadratic formula to determine \(x\). We can identify that \(\textcolor{red}{a = 4}\), \(\textcolor{green}{b = 12}\) and \(\textcolor{blue}{c = -616}\):
\(x = \cfrac{-\textcolor{green}{b} +- \sqrt{\textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c}}}{2\textcolor{red}{a}}\)
\(x = \cfrac{-(\textcolor{green}{12}) +- \sqrt{(\textcolor{green}{12})^2 - 4(\textcolor{red}{4})(\textcolor{blue}{-616})}}{2(\textcolor{red}{4})}\)
\(x = \cfrac{-4 \pm \sqrt{144 + 9856}}{8}\)
\(x = \cfrac{-12 \pm \sqrt{10000}}{8}\)
Using \(\pm\) we can determine \(x\) either by adding or subtracting the square root value:
We can determine the first potential height by adding the square root value:
\(x_1 = \cfrac{-12 + 100}{8}\)
\(x_1 = \cfrac{88}{8}\)
\(x_1 = 11\)
We can determine the second potential height by adding the square root value:
\(x_2 = \cfrac{-12 - 100}{8}\)
\(x_2 = \cfrac{-112}{8}\)
\(x_2 = -14\)
We can determine that \(\boldsymbol{\textcolor{red}{x = 11}}\) where the consecutive numbers are \(\boldsymbol{\textcolor{red}{11, 12, 13, 14}}\).
We can also determine that \(\boldsymbol{\textcolor{blue}{x = -14}}\) where the consecutive numbers are \(\boldsymbol{\textcolor{blue}{-14, -13, -12, -11}}\).
A model rocket is launched from the deck that is \(15\;[\text{m}]\) high, with an initial speed of \(100\;[\text{m/s}]\).
i. In order to model an equation for the launch of the rocket, we can use the formula for projectile motion and plug all the given values into it:
\(h = -4.9t² + Vt + H\)
\(h = -4.9t² + 100t + 15\)
Therefore, we can determine that the equation used to model this scenario is \(h = -4.9t² + 100t + 15\).
ii. All we need to do to determine the height of the rocket after \(2\;[\text{s}]\) is plug substitute \(2\) for \(t\) in the equation:
\(h = -4.9t² + 100t + 15\)
\(h = -4.9(2)² + 100(2) + 15\)
\(h = -4.9(4) + 200 + 15\)
\(h = -19.6 + 200 + 15\)
\(h = 195.4\;[\text{m}]\)
Therefore, we can determine that the rocket's height after \(2 \; [\text{s}]\) is \(\textbf{195.4[m]}\).
iii. In order to determine the maximum height, we need the vertex. We can identify the vertex by Completing the Square:
\(h = -4.9(t² - 20.4t) + 15\)
\(\left(\cfrac{b}{2}\right)² = \left(\cfrac{-20.4}{2}\right)² = (-10.2)² = 104.04\)
\(h = -4.9(t² - 20.4t + 104.04 - 104.04) + 15\)
\(h = -4.9(t² - 20.4t + 104.04) -4.9(-104.04) + 15\)
\(h = -4.9(t² - 10.2t - 10.2t + 104.04) + 509.796 + 15\)
\(h = -4.9((t(t - 10.2) - 10.2(t - 10.2)) + 524.796\)
\(h = -4.9((t -10.2)²) + 524.796\)
Therefore, we can determine that the maximum height of the rocket is \(\approx \textbf{ 525 [m]}\).
iv. By determining the vertex of the quadratic in the previous question, we can determine that it took \(\textbf{10.2 [s]}\) for the rocket to reach maximum height.