Answer:
border area: 100 m²
Step-by-step explanation:
The area of the pool border will be the overall area of the pool and boder, less the area of the pool. The area of each rectangle is the product of its length and width. The length and width of each can be found using the Pythagorean theorem.
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<h3>Pythagorean Triple</h3>The triple (3, 4, 5) is called a Pythagorean triple, because ...
3² +4² = 5²
That is, a triangle with sides of lengths 3, 4, and 5 will be a right triangle with a hypotenuse of length 5.
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<h3>Application to Pool/Border Dimensions</h3>We notice that the sides of the rectangles in the figure have a rise/run of ±3/4 or ±4/3. In each case, the side of the rectangle is the hypotenuse of a triangle with dimensions a multiple of 3 or 4 meters. Hence the hypotenuse (rectangle side length) is the corresponding multiple of 5 meters. The attachment illustrates this point, showing a triangle with sides that are (6, 8, 10), 2 times the basic triple.
The pool is 5 m by 10 m, for an area of ...
A = LW = (10 m)(5 m) = 50 m²
The overall area is 10 m by 15 m, for an area of ...
A = (15 m)(10 m) = 150 m²
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<h3>Border Area</h3>The pool's border is the difference between the overall area and the pool area:
150 m² -50 m² = 100 m² . . . . area of pool's border
Answer:
0.25ab
Step-by-step explanation:
Data provided in the question:
f(x) = xa(1−x)b, 0≤x≤1
or
f(x) = ab(x−x²)
for point of maxima and minima put f'(x) = 0
Thus,
f'(x) = ab(1 - 2x) = 0
or
1 - 2x = 0
or
x = = 0.5
Now,
to check the condition of maxima or minima
f''(x) = ab(0 - 2) = -2ab
since,
f''(x) < 0
therefore,
x = 0.5 is point of maxima
and the maximum value at x = 0.5 of the function is
f(0.5) = ab(0.5 - 0.5²)
= ab(0.25)
= 0.25ab