volume?
Area=1/2 b*h but 20^2=b^2+h^2 or
h=sqrt (400-b^2)
area= 1/2 * sqrt( )*b= 1/2 b*sqrt ( )
take the derivative, set darea/db =0, ans solve for b.
Area=1/2 b*h but 20^2=b^2+h^2 or
h=sqrt (400-b^2)
area= 1/2 * sqrt( )*b= 1/2 b*sqrt ( )
take the derivative, set darea/db =0, ans solve for b.
and the maximum area is thus 200 cm^2
A=1/2 b x h
b^2 + h^2 = 20^2 → h = sqrt (400 - b^2)
Substitute h to A
A = (1/2)(b)sqrt(400-b^2)
Derive both sides and let dA/dt = 0
Find b
You can get b= 10sqrt(2)
Subs to h then you can get h=10sqrt(2)
A= 1/2 bxh
A=200/2
A=200 sq.cm
A= 100 sq cm
In a right triangle, the hypotenuse is opposite the right angle and is the longest side. Let's assume that the two legs of the triangle have lengths a and b (where a < b).
According to the Pythagorean theorem, which relates the lengths of the sides of a right triangle, the sum of the squares of the two legs equals the square of the hypotenuse. In this case, we have:
a^2 + b^2 = 20^2
To find the maximum possible volume, we need to maximize the product of the two legs, a and b. In other words, we are looking for the maximum possible value of the function V = a * b.
To solve the problem, we can express one of the variables in terms of the other using the Pythagorean theorem equation and then substitute it into the volume equation. Let's solve for b in terms of a:
b^2 = 20^2 - a^2
b = sqrt(400 - a^2)
Now we can substitute this expression for b into the volume equation:
V = a * sqrt(400 - a^2)
To find the maximum volume, we need to find the value of a that maximizes this function. We can do this by taking the derivative of the volume function with respect to a, setting it equal to zero, and then solving for a. However, since taking the derivative can be complicated, we can use a graphing calculator or computer software to find the maximum.
By plotting the volume function, we can see that it reaches its maximum value when a is approximately equal to 14.142 cm. Plugging this value of a back into the volume equation, we can find the maximum volume.