Maximal feild question

• transgalactic
In summary, the conversation discusses how to calculate the maximal space of an isosceles trapeze inside an ellipse. The method involves finding the two x-values that correspond to a given y-value, which can then be used to determine the length of the smaller base. The area of the trapeze can then be written as a function of y and maximized to find the maximal space, which is equal to 3^0.5(a*b). The conversation also mentions using integrals and the formula of the trapeze function, but these methods were not successful.

transgalactic

i added a photo of the situation

inside an ellipse (its formula x^2 /a^2 + y^2 /b^2 =1)
we block an isosceles trapeze .
the large base of the trapeze is on the X axes.

show that the maximal space of the trapeze is 3^0.5( a*b)?

i tried to get the field of the trapeze by signing the height by X

but if i connect it to the ellipse formula for as x-b

i need another variable for the small base of the trapeze.

how can i solve it?

i tried by integrals but i don't know the formula of the trapeze fonction

what to do?
how do i prove that the maximal space of the trapeze is 3^0.5( a*b)?

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Since the height is perpendicular to the x-axis, call it y instead!

What two x-values correspond to y? Their difference is the length of the smaller base.

there is no x vaules that are connected to the length of
the small base

we can say that the length of the big base is 2a
but in order to find the little base or the height of the trapeze
there begins the problem
notice that the answer includes only the parameters of the
ellipse fuction??

Did you read my first response? Your last post gives no indication that you have!

Solve
$$\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$$
for x as a function of y. You will get two values of x. There difference is the length of the smaller base. With that you can write the area of the trapezoid as a function of y and then maximize that function.

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thank u very much

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