Solution 1.3:4

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If we call the ''x''-coordinate of the point <math>P</math> <math>x</math>, then its ''y''-coordinate is <math>1-x^{2}</math>, because <math>P</math> lies on the curve <math>y=1-x^{2}</math>.
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The area of the rectangle is then given by
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{{Displayed math||<math>A(x) = \text{(base)}\cdot\text{(height)} = x\cdot (1-x^2)</math>}}
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and we will try to choose <math>x</math> so that this area function is maximised.
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To begin with, we note that, because <math>P</math> should lie in the first quadrant, <math>x\ge 0</math> and also <math>y=1-x^2\ge 0</math>, i.e. <math>x\le 1</math>. We should therefore look for the maximum of <math>A(x)</math> when <math>0\le x\le 1\,</math>.
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There are three types of points which can maximise the area function:
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# critical points,
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# points where the function is not differentiable,
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# endpoints of the region of definition.
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The function <math>A(x) = x(1-x^2)</math> is differentiable everywhere, so item 2 does not apply. In addition, <math>A(0) = A(1) = 0\,</math>, so the endpoints in item 3 cannot be maximum points (but rather the opposite, i.e. minimum points).
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We must therefore conclude that the maximum area is a critical point. We differentiate
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{{Displayed math||<math>A'(x) = 1\cdot (1-x^2) + x\cdot (-2x) = 1-3x^2\,,</math>}}
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and the condition that the derivative should be zero gives that <math>x=\pm 1/\!\sqrt{3}</math>; however, it is only <math>x=1/\!\sqrt{3}</math> which satisfies <math>0\le x\le 1</math>.
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At the critical point, the second derivative <math>A''(x)=-6x</math> has the value
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{{Displayed math||<math>A''\bigl( 1/\!\sqrt{3}\bigr) = -6\cdot\frac{1}{\sqrt{3}} < 0\,,</math>}}
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which shows that <math>x=1/\!\sqrt{3}</math> is a local maximum.
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The answer is that the point <math>P</math> should be chosen so that
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{{Displayed math||<math>P = \Bigl(\frac{1}{\sqrt{3}}, 1-\Bigl(\frac{1}{\sqrt{3}} \Bigr)^2\, \Bigr) = \Bigl(\frac{1}{\sqrt{3}}, \frac{2}{3} \Bigr)\,\textrm{.}</math>}}

Current revision

If we call the x-coordinate of the point \displaystyle P \displaystyle x, then its y-coordinate is \displaystyle 1-x^{2}, because \displaystyle P lies on the curve \displaystyle y=1-x^{2}.

The area of the rectangle is then given by

\displaystyle A(x) = \text{(base)}\cdot\text{(height)} = x\cdot (1-x^2)

and we will try to choose \displaystyle x so that this area function is maximised.

To begin with, we note that, because \displaystyle P should lie in the first quadrant, \displaystyle x\ge 0 and also \displaystyle y=1-x^2\ge 0, i.e. \displaystyle x\le 1. We should therefore look for the maximum of \displaystyle A(x) when \displaystyle 0\le x\le 1\,.

There are three types of points which can maximise the area function:

  1. critical points,
  2. points where the function is not differentiable,
  3. endpoints of the region of definition.

The function \displaystyle A(x) = x(1-x^2) is differentiable everywhere, so item 2 does not apply. In addition, \displaystyle A(0) = A(1) = 0\,, so the endpoints in item 3 cannot be maximum points (but rather the opposite, i.e. minimum points).

We must therefore conclude that the maximum area is a critical point. We differentiate

\displaystyle A'(x) = 1\cdot (1-x^2) + x\cdot (-2x) = 1-3x^2\,,

and the condition that the derivative should be zero gives that \displaystyle x=\pm 1/\!\sqrt{3}; however, it is only \displaystyle x=1/\!\sqrt{3} which satisfies \displaystyle 0\le x\le 1.

At the critical point, the second derivative \displaystyle A''(x)=-6x has the value

\displaystyle A''\bigl( 1/\!\sqrt{3}\bigr) = -6\cdot\frac{1}{\sqrt{3}} < 0\,,

which shows that \displaystyle x=1/\!\sqrt{3} is a local maximum.

The answer is that the point \displaystyle P should be chosen so that

\displaystyle P = \Bigl(\frac{1}{\sqrt{3}}, 1-\Bigl(\frac{1}{\sqrt{3}} \Bigr)^2\, \Bigr) = \Bigl(\frac{1}{\sqrt{3}}, \frac{2}{3} \Bigr)\,\textrm{.}
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