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urugvai
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Consider squares inscribed in different isosceles triangles with sides equal to 1. (One side of the square lies on the bottom.) Find the side of the largest squareView attachment 6546
Thank you very much, but now the solution implies the derivation of the formula for the dependence of the area on some value (which should be chosen), the derivative, critical points, and so on. And the proposed answer introduced me into to a grinding halt.Evgeny.Makarov said:Hi, and welcome to the forum!
You can do this either with or without trigonometry. But in either case I have not found a formula for the largest square side, only an approximate value.
Let the $AC=b$ and $BH=h$. Then the area of $\triangle ABC$ is $(1/2)bh$. It also equals the sum of areas of $\triangle KBL$, the area of the square and the combined area of $\triangle AKN$ and $\triangle CLM$. Let the square side be $x$. Then the combined area of $\triangle AKN$ and $\triangle CLM$ is $(1/2)x(b-x)$. Can you express the area of the remaining figures and write the equation?
For the future, please read the http://mathhelpboards.com/rules/, especially rule #11 (click the Expand button on top).
We may choose $b$ as the parameter. It varies between 0 and 2. Alternatively, we can choose $h$, which varies between 0 and 1.urugvai said:now the solution implies the derivation of the formula for the dependence of the area on some value (which should be chosen)
Yes, this has to be done, but you should show your work or describe your difficulties.urugvai said:the derivative, critical points, and so on.
This value seems correct, but deriving it analytically will require some work.urugvai said:And the proposed answer introduced me into to a grinding halt.
MarkFL said:\(\displaystyle \frac{x}{\sqrt{1-b^2}}+\frac{y}{b}=1\)
urugvai said:Thanks, but there was a difficulty, is this the sum of vectors?
MarkFL said:I used the two-intercept form of a line, and the fact that the distance between the two intercepts must be 1. :D
urugvai said:The temperature of the brain in the head only increases :(
MarkFL said:A line having $x$-intercept at $(a,0)$ and $y$-intercept at $(0,b)$ will be given by:
\(\displaystyle \frac{x}{a}+\frac{y}{b}=1\)
Now, if the distance between these intercepts is 1, then we may write:
\(\displaystyle \sqrt{(a-0)^2+(0-b)^2}=1\)
This implies:
\(\displaystyle a^2=1-b^2\)
If we assume $0<a$, then we have:
\(\displaystyle a=\sqrt{1-b^2}\)And so we obtain the line:
\(\displaystyle \frac{x}{\sqrt{1-b^2}}+\frac{y}{b}=1\)
urugvai said:Thank you for explaining this in more detail. Sorry that you need to spend so much time explaining
urugvai said:Two more questions
X is the half of the side of the square
Y is the side of the square?
Extrema refers to the maximum or minimum values of a function. It can be either a local extrema, which is the highest or lowest point in a small interval, or a global extrema, which is the highest or lowest point in the entire domain of the function.
To determine the extrema of a function, one must first take the derivative of the function and set it equal to zero. Then, solve for the variable in the equation to find the critical points. Next, plug in the critical points into the original function to determine whether they are maximum or minimum values.
Finding extrema in a function can help us understand the behavior of the function and its graph. It can also help us optimize the function for a specific purpose, such as finding the maximum profit or minimum cost.
Yes, a function can have multiple extrema, both local and global. This means that there can be more than one highest or lowest point in a small interval or in the entire domain of the function.
Yes, the most common method for finding extrema in a function is using the first derivative test. However, there are also other methods such as the second derivative test and the closed interval method. The choice of method depends on the complexity of the function and the specific problem at hand.