<span>The steps for using a compass and straightedge to construct an equilateral triangle are as follows:
1. </span><span>Construct a circle with point E as the center and a circle with point F as the center with each circle having radius EF
2. </span><span>Construct line EF
3. </span><span>Construct a point at one of the two intersection points of circle E and circle F and label this point G
4. </span><span>Construct line EG and line FG.</span>
The absolute minimum = -2√2.
The absolute maximum= 4.5
Consider f(t)=t√9-t on the interval (1,3].
Find the critical points: Find f'(t)=0.
f"(t) = 0
√9-t² d/dt t + t d/dt √9-t²=0
√9-t² + t/2√9-t² (-2t)=0
9-t²-t²/√9-t²=0
9-2t²=0
9=2t², t²=9/2, t=±3/√2
since -3/√2∉ (1,3].
Therefore, the critical point in the interval (1,3] is t= 3/√2.
Find the value of the function at t=1, 3/√2,3 to find the absolute maximum and minimum.
f(-1)=-1√9-1²
= -√8 , =-2√2
f(3/√2)= 3/√2 √9-(3/√2)²
= 3/√2 √9-9/2
=3/√2 √9/2
=9/2 = 4.5
f(3)= 3√9-3²
= 3(0)
=0
The absolute maximum is 4.5 and the absolute minimum is -2√2.
The absolute maximum point is the point at which the function reaches the maximum possible value. Similarly, the absolute minimum point is the point at which the function takes the smallest possible value.
A relative maximum or minimum occurs at an inflection point on the curve. The absolute minimum and maximum values are the corresponding values over the full range of the function. That is, the absolute minimum and maximum values are bounded by the function's domain.
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The rate at which the tank is filled up, F, and emptied, E, are respectively:
Since V is the full volume of the tank, the time required to fill the tank if both pipes are open is given by:
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