Consider a right-angled triangle with a hypotenuse equal to 8 cm. At the request of the task, we find the length of each leg if the area of the triangle should be the largest.
The legs and hypotenuse of this right-angled triangle will be denoted, respectively, by a, b and c. Then, according to the conditions of the task, c = 8, and the legs a and b are unknown (here and further in the calculations, until the final result, we will omit the unit of length cm).
We will use the Pythagorean theorem, the formula of which for our task can be formalized as the equality c² = a² + b² or, after substituting 8 instead of the hypotenuse, 8² = a² + b². Consider the last equality as an equation with respect to one leg (for example, with respect to leg b) and solve it. Then, we have: b² = 64 – a², whence b = ± √ (64 – a²). Since the leg of a right-angled triangle cannot be negative, we get: b = √ (64 – a²).
As you know, the area S of a right-angled triangle can be calculated by the formula S = ½ * a * b, where a and b are the legs of a right-angled triangle. Substituting into this formula the last expression for b from the previous paragraph. Then, we get the following function (from the argument a) S (a) = ½ * a * √ (64 – a²). Thus, the task conditions were reduced to finding the largest value of the function S (a) = ½ * a * √ (64 – a²) in the interval (0; 8).
To solve this problem, we will apply the methods of differential and integral calculus. For this purpose, we calculate the derivative of the function S (a) = ½ * a * √ (64 – a²). We have: S Ꞌ (a) = (½ * a * √ (64 – a²)) Ꞌ = ½ * (aꞋ * √ (64 – a²) + a * (√ (64 – a²)) Ꞌ) = ½ * ( √ (64 – a²) + a * ½ * (-2 * a / √ (64 – a²))) = ½ * ((√ (64 – a²)) ² – a * a) / √ (64 – a²) ) = ½ * (64 – a² – a²) / √ (64 – a²) = (32 – a²) / √ (64 – a²).
Equating to zero the derivative (S Ꞌ (a) = 0), we define stationary points (if, of course, there are any) of the function S (a). We have: (32 – a²) / √ (64 – a²) = 0. Since we consider the function S (a) in the interval (0; 8), the denominator of the fraction on the left side of the equation cannot vanish. Therefore, this equation can rewrite as 32 – a² = This is an incomplete quadratic equation with respect to a, has two different roots a = ± √ (32) = ± 4√ (2). Obviously, the root a = -4√ (2) is a side root. Hence, we choose the root a = 4√ (2).
Consider two intervals (0; 4√ (2)) and (4√ (2); 8). It is easy to verify that S Ꞌ (a)> 0 for a ∈ (0; 4√ (2)) and S Ꞌ (a) <0 for a ∈ (4√ (2); This means that the function S (a) at the point a = 4√ (2) takes the maximum (largest) value. Now we can easily calculate the other leg of a right triangle: b = √ (64 – a²) = √ (64 – (4√ (2)) ²) = √ (64 – 32) = √ (32) = 4√ (2) We conclude: With equal legs of 4√ (2) cm each, the area of the triangle will be the largest.
Answer: With equal legs of 4√ (2) cm each, the area of the triangle will be the largest.
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