Question

A right triangle is formed in the first quadrant by the \(x\) - and \(y\) -axes and a line through the point \((1,2)\) (see figure). (a) Write the length \(L\) of the hypotenuse as a function of \(x\). (b) Use a graphing utility to approximate \(x\) graphically such that the length of the hypotenuse is a minimum. (c) Find the vertices of the triangle such that its area is a minimum.

Short answer

The length \(L\) of the hypotenuse is a function of \(x\) given by \(L = \sqrt{(2-x)^2 + x^2}\). The minimum length of hypotenuse and vertices of the triangle for the minimum area can be obtained by differentiating the \(L\) and \(A\) functions respectively and setting them equal to zero.

Step-by-step solution

Formulate Hypotenuse Length Function

The length of the hypotenuse, \(L\), in the right triangle can be expressed using the Pythagorean theorem. The coordinates of the point where the hypotenuse intersects with the \(x\)-axis is \((x,0)\), and with the \(y\)-axis is \((0,2-x)\), due to the line equation \(y=2-x\). So, we have \(L = \sqrt{(2-x)^2 + x^2}\).

Approximate Minimum Hypotenuse Length

To find the minimum length of hypotenuse, one needs to plot the function \(L = \sqrt{(2-x)^2 + x^2}\) and find the minimum point in the graph. For an accurate result, use a graphing calculator or a mathematics software.

Find Triangle Vertices with Minimum Area

The area \(A\) of the right triangle can be calculated as \(A =\frac{1}{2} * \text{base} * \text{height} =\frac{1}{2} * x * (2-x)\). To find the value of \(x\), differentiate this function and solve \(A'=0\). Using the value of \(x\), the vertices of the triangle can be found. The vertices are \((0, 0)\), \((x, 0)\), and \((0, 2-x)\).

Key concepts

Pythagorean theorem

The Pythagorean theorem is a fundamental principle in geometry that relates the lengths of the sides in a right-angled triangle. More specifically, it states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Mathematically, if we denote the hypotenuse as 'c' and the other two sides as 'a' and 'b', the Pythagorean theorem can be expressed as: \[ c^2 = a^2 + b^2 \]
In the context of the exercise, we utilize this theorem to write the length of the hypotenuse, \(L\), as a function of the horizontal distance, \(x\), using the coordinates provided by the point \((1,2)\) and the axes of the coordinate system. The resulting equation \(L = \sqrt{(2-x)^2 + x^2}\) emerges from the relationship that the theorem describes.

Graphing utility

Graphing utilities, such as calculators or software programs, offer valuable visual representations of mathematical functions. They allow us to plot equations and observe their behavior over a range of values. By graphing the hypotenuse length function \(L\), we can visually identify the point where the graph reaches its lowest value—indicating the minimum length of the hypotenuse.

Using a graphing utility not only aids in identifying this minimum point but also provides a practical understanding of function behavior. For instance, in Step 2 of the solution, we seek to find the exact value of \(x\) that minimizes the function \(L(x)\) within the first quadrant. Graphing utilities enable us to accomplish this task visually and interactively, which can be especially useful when analytic methods are challenging to apply.

Minimum hypotenuse length

The minimum hypotenuse length in a geometric problem is often a point of optimization, which means finding the smallest possible length of the hypotenuse that satisfies the given conditions. In this exercise, we need to identify the x-coordinate at which the hypotenuse length reaches its minimum. By using the function we derived from the Pythagorean theorem, \(L = \sqrt{(2-x)^2 + x^2}\), we can investigate this optimization problem.

While a graphing utility can provide an approximate answer graphically, further mathematical techniques, such as differentiation, allow us to find a precise value. Identifying the minimum length involves setting the derivative of the length function to zero and solving for \(x\), which represents an important technique not only in geometry but in various optimization problems across mathematics and related fields.

Triangle area minimization

Minimizing the area of a triangle while maintaining specific constraints represents an optimization problem. In the case of the given right triangle, we want to reduce its area to the smallest possible value based on its relationship with the axes and a fixed point. The general formula for the area of a triangle is \(A = \frac{1}{2} * \text{base} * \text{height}\).

Using the vertices provided by the problem, namely \((0, 0)\), \((x, 0)\), and \((0, 2-x)\), we express the area as a function of \(x\): \(A(x) = \frac{1}{2} * x * (2-x)\). Once we have this function, we can apply differentiation to find the x-coordinate at which this function is minimized, ultimately leading to the triangle vertices that correspond to the minimum area.

Differentiation

Differentiation is a technique in calculus used to find the rate at which a function changes at any given point. It is the process of computing a derivative, which measures how a function's value will change as its input changes. In optimization problems such as finding the minimum length of the hypotenuse or the area of a triangle, the derivative helps us determine where these functions reach their minimum values.

In our exercise, we differentiate the area function \(A(x)\) with respect to \(x\), and solve the equation \(A'(x) = 0\) to find the critical points. These critical points are candidates for local minimums or maximums of the function. Once we have identified these points, we can test them to determine whether they correspond to minimum values for the problem we are trying to solve. This mathematical process of differentiation proves essential in solving many real-world problems that involve optimization.