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Magazine Chapter 5: Problem 7
Suppose that \(a\) and \(b\) are the side lengths in a right triangle whose hypotenuse is \(10 \mathrm{~cm}\) long. Show that the area of the triangle is largest when \(a=b\).
Short Answer
Expert verified
The area of the triangle is largest when both sides are \(\sqrt{50} \) cm, meaning \(a = b\).
Step by step solution
01
Understand the Problem
In the right triangle, we are given that the hypotenuse is fixed at 10 cm. We need to find the values of sides \(a\) and \(b\) such that the area of the triangle is maximized, specifically when \(a=b\).
02
Express the Area of Triangle
The area \(A\) of a right triangle is given by \(A = \frac{1}{2}ab\). We need to express this area in terms of \(a\) and \(b\).
03
Use Pythagorean Theorem
Since it is a right triangle, the Pythagorean theorem gives \(a^2 + b^2 = 10^2 = 100\). This relates \(a\) and \(b\).
04
Substitute for One Variable
Assume \(b = x\), thus from the Pythagorean theorem \(a^2 = 100 - x^2\). Therefore, \(a = \sqrt{100 - x^2}\).
05
Express Area in Terms of a Single Variable
Substitute \(a = \sqrt{100 - x^2}\) into the area formula: \[A = \frac{1}{2} x \sqrt{100 - x^2} \]
06
Maximize the Area Function
To find the maximum area, we take the derivative of \(A\) with respect to \(x\) and set it to zero: \[ \frac{dA}{dx} = \frac{1}{2} \left( \sqrt{100 - x^2} - \frac{x^2}{\sqrt{100 - x^2}} \right) = 0 \]Solve for \(x\).
07
Solve the Equation
Equating it to zero, solve \[ \sqrt{100 - x^2} = \frac{x^2}{\sqrt{100 - x^2}} \] which simplifies to \[ 100 - x^2 = x^2 \] Thus, \[ 2x^2 = 100 \] Thus, \[ x^2 = 50 \] This gives \(x = \sqrt{50}\).
08
Check the Condition a = b
Since \(x = \sqrt{50}\) and from Step 4, \(a = \sqrt{100 - x^2} = \sqrt{50}\). Hence, \(a = b = \sqrt{50}\). The maximum area is achieved when \(a=b\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Right Triangle
A right triangle is a type of triangle where one of the angles measures exactly 90 degrees. This makes it unique and very useful for various calculations in geometry. In a right triangle, the side opposite the right angle is always the longest side and is called the hypotenuse. The other two sides are referred to as the "legs" or "catheti" of the triangle.
When solving problems involving right triangles, a common task is to find the lengths of these sides or utilize them for further calculations, like finding the area or applying trigonometric ratios. Since the hypotenuse is the longest side in a right triangle:
When solving problems involving right triangles, a common task is to find the lengths of these sides or utilize them for further calculations, like finding the area or applying trigonometric ratios. Since the hypotenuse is the longest side in a right triangle:
- The hypotenuse always serves as a crucial reference point, especially in optimization problems as in our case where we look for side lengths that maximize the area of the triangle.
- Main properties, like the Pythagorean Theorem and trigonometry, largely depend on the presence of a right angle and the properties of the hypotenuse.
Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle in geometry, especially related to right triangles. 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. This can be expressed with the formula:
\[ a^2 + b^2 = c^2 \]
Where:
For our problem, the hypotenuse is 10 cm, and using the Pythagorean Theorem, we got:
\[ a^2 + b^2 = 10^2 = 100 \]
This relationship allows us to express one side in terms of the other, leading to secondary functions and optimization.
\[ a^2 + b^2 = c^2 \]
Where:
- \(a\) and \(b\) are the lengths of the legs,
- \(c\) is the length of the hypotenuse.
For our problem, the hypotenuse is 10 cm, and using the Pythagorean Theorem, we got:
\[ a^2 + b^2 = 10^2 = 100 \]
This relationship allows us to express one side in terms of the other, leading to secondary functions and optimization.
Derivative
In calculus, the derivative of a function represents how a function's output changes as its input changes. It is a fundamental tool for finding the maximum or minimum values of functions, which is an essential aspect of optimization problems. In this problem, we use derivatives to find the point at which the area of the triangle is maximized.
To find the maximum area of the triangle, we express the area function in terms of one variable derived from the condition set by the Pythagorean Theorem. Then, we take the derivative of this area function and set it to zero to find critical points:
\[ \frac{dA}{dx} = \frac{1}{2} \left( \sqrt{100 - x^2} - \frac{x^2}{\sqrt{100 - x^2}} \right) \]
This application illustrates how calculus helps in efficiently solving real-world problems, like optimizing function results to achieve desired outcomes, such as maximizing the area of a triangle given specific conditions.
To find the maximum area of the triangle, we express the area function in terms of one variable derived from the condition set by the Pythagorean Theorem. Then, we take the derivative of this area function and set it to zero to find critical points:
\[ \frac{dA}{dx} = \frac{1}{2} \left( \sqrt{100 - x^2} - \frac{x^2}{\sqrt{100 - x^2}} \right) \]
This application illustrates how calculus helps in efficiently solving real-world problems, like optimizing function results to achieve desired outcomes, such as maximizing the area of a triangle given specific conditions.
