How Do You Maximize the Volume of a Prism with an Equilateral Triangle Base?

[rachael]

3 The cross-section of a solid prism is an equilateral triangle, The sum of the lengths of the edges of the prsim is 18 cm.
a. find the exact side length of the triangle in cm so that the volume is maximised.
b. Find the maximum volume in cm^3, correct to 2dp

i can't seem to find the correct answer for this question
could anyone please help me?
thank you

 

[pavadrin]

3 The cross-section of a solid prism is an equilateral triangle, The sum of the lengths of the edges of the prsim is 18 cm.
a. find the exact side length of the triangle in cm so that the volume is maximised.
b. Find the maximum volume in cm^3, correct to 2dp

i can't seem to find the correct answer for this question
could anyone please help me?
thank you

wheres ur working?

 

[Architeuthis Dux]

I would first like to commend you for your efforts in attempting to solve this problem. Differentiation is a powerful tool in mathematics and can be applied to various real-world scenarios. In this case, we are dealing with a solid prism with an equilateral triangle cross-section and a given sum of edge lengths. Let's break down the problem and use differentiation to find the maximum volume.

a. To find the exact side length of the triangle, we can use the formula for the volume of a prism, which is V = Bh, where B is the area of the base and h is the height. In this case, the base is an equilateral triangle, so we can use the formula for the area of an equilateral triangle, which is A = √3/4 x s^2, where s is the side length. We also know that the sum of the edge lengths is 18 cm, so we can write the equation s + s + s = 18, or 3s = 18. Solving for s, we get s = 6 cm. Therefore, the exact side length of the equilateral triangle is 6 cm.

b. To find the maximum volume, we can use the first derivative test. We know that the volume is a function of the side length, so we can write V(s) = √3/4 x s^2 x h. Using the formula for the height of an equilateral triangle, which is h = √3/2 x s, we can rewrite the volume function as V(s) = √3/4 x s^3. To find the maximum volume, we need to find the critical points of this function. Taking the first derivative, we get V'(s) = 3√3/4 x s^2. Setting this equal to 0 and solving for s, we get s = 0. Therefore, the only critical point is s = 0.

To determine if this is a maximum or minimum, we can use the second derivative test. Taking the second derivative, we get V''(s) = 3√3/2 x s. Plugging in s = 0, we get V''(0) = 0. Since the second derivative is 0, we cannot determine if this is a maximum or minimum using this test. However, we can use the endpoint test since we know that s