I think this question is confusing.

There can only be 1 fixed value of r which can form the figure given that the total length is 80 cm.

That value of r is 80/(2×sqrt(2) +pi).

Since r is fixed, it makes no sense to talk about maximum or minimum area.

ii)

A = ½πr² + ⅛(80 - πr)²

dA/dr = πr + ¼(80 - πr)(-π)

= πr - 20π + ¼π²r

= π(1+ ¼π)r - 20π

When dA/dr = 0,

π(1+ ¼π)r - 20π = 0

(1+ ¼π)r - 20 = 0

(1+ ¼π)r = 20

r = 20/(1 + ¼π)


iii)


Second derivative method :

dA/dr = π(1+ ¼π)r - 20π

So d²A/dr² = π(1 + ¼π) = π + ¼π² > 0

(Minimum)

J.

I disagree.

That expression for A in terms of r, if treated just as a function of r is amenable to the usual methods of Calculus.

But in this case, to form that particular figures, there can be only 1 value of r.

Yes, the question is flawed. I do not dispute your explanation.

But the student has asked for the working to solve part iii) so I have provided it.

The problem with this question should be raised with the teachers or relevant publishers.

Edit : mistake in working. Amended

Fpr part (iii) if i wan to use the 1st Derivative test. How do I do it??

You'll have to find the values of dA/dr when r is bigger than and also when r is smaller than 20/(1 + ¼π).

Use values as close to 20/(1 + ¼π) as possible.

You will find that dA/dr is negative for the value smaller than 20/(1 + ¼π) and positive for the value bigger than 20/(1 + ¼π)

This means the curve is upward sloping (U -shaped) and therefore it's a minimum.

Edit : corrected dA/dr

Visual representation :

r

∥ 20/(1+¼π)- ∥ 20/(1 + ¼π) ∥ 20/(1 + ¼π)+ ∥

dA/dr

∥____ \________∥_____‾______∥ _______/______∥

Since r = 20/(1 + ¼π) ≈ 11.2019830702 ,

You can try with values such as r = 11.19 and r = 11.21

Thank u so much J.

Welcome