An isosceles triangle has its vertex at the origin and its bas...

Question

An isosceles triangle has its vertex at the origin and its base parallel to the x-axis with the vertices above the axis on the curve y = 27 - x². Find the largest area the triangle can have.

Accepted Answer

A parabolic arch has its base on the x-axis and its highest point at Y equals 64 minus 4 X2. If a rectangle is inscribed in the arch with one side on the X axis and opposite vertices on the curve, what is the largest area the rectangle can have? Now, I provided a little diagram just to show what this looks like. We have a rectangle Underneath a parabola and the vertices of the rectangle on the parabola. Now the area Of a rectangle We know his wits. Multiplied by height. Let's find functions for each of these attributes. For the width, we noticed that since we have a negative X and positive X for our vertices. Our width will just be 2 X, which is the sum. Of the two side lengths. The height is given by our function 64 minus 4 X squared. So our area then will be 128 X. Minus 8 execute. Now we need to find the maximum area. To find the maximum area, we will find the critical point. We will say A is equal to 0. We have 128. Minus 24 X squared by the power rule, that equals 0. Now, let's find the X value where we have a maximum by solving for X. We have 24 X2 equals 128. Divide both sides by 24. To get X2 equals 16 divided by 3 when simplified. Now, if we take the square root of both sides, we get X equals plus or minus the square root of 16 divided by 3, which simplifies C+ or minus 4 square of 3 divided by 3. Now, since you want maximum area. And the width of the rectangle should be a positive value. We will only take the positive form of 4 scoops of 3 divided by 3. Now, let's confirm that this is indeed the maximum by checking the 2nd derivative. We just need to see if the second derivative is negative at our point. So our second derivative. will be 48 X. If you plug in our acrylical point. 48, multiplied by 4 squares of 3 divided by 3. We get -642 of 3. This is negative, so we can confirm that this is a maximum. Now, the maximum area, we will take our X value and plug it back into our equation. So we have a max area. Of 128. Multiplied by 4 squares of 3 divided by 3. Minus 8 multiplied by 4 square to 3 divided by 3 Q. Now we can just simplify this. We will get 128, multiplied by 4 square of 3, divided by 3. Which gives us 512 spirits of 3 divided by 3. 84 sorts of 3 divided by 3 cubed. We can simplify. To get 512 squares of 3 divided by 9. We can now simplify this once more, begging the common denominator. To get us 1024 squares of 3. Divided by 9. Since this is area, we will use square units. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

An isosceles triangle has its vertex at the origin and its base parallel to the x-axis with the vertices above the axis on the curve y = 27 - x². Find the largest area the triangle can have.

Key Concepts

Area of a Triangle

Quadratic Functions

Optimization

Watch next

An isosceles triangle has its vertex at the origin and its base parallel to the x-axis with the vertices above the axis on the curve y = 27 - x2. Find the largest area the triangle can have.

Key Concepts

Area of a Triangle

Quadratic Functions

Optimization

Watch next

An isosceles triangle has its vertex at the origin and its base parallel to the x-axis with the vertices above the axis on the curve y = 27 - x². Find the largest area the triangle can have.