Optimisation with Geometric Constraints | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

A cross-section can supply the constraint

A cylinder fits inside a sphere of radius 5 cm, with both circular rims touching the sphere. Let its radius be $r$ cm and its full height be $2h$ cm. Cut vertically through the common centre.

The full cylinder height is 2h. The right triangle uses only the upper half, h.

The sphere centre is halfway up the cylinder. The right triangle has vertical side $h$ , horizontal side $r$ and hypotenuse 5. Pythagoras gives $r^2+h^2=25$ . Keep $r^2=25-h^2$ ; the volume needs the square, so taking a square root would add work.

\begin{gathered}V=\pi r^2(2h)\\=50\pi h-2\pi h^3\\0<h<5\\V'=50\pi-6\pi h^2=0\\h=\frac5{\sqrt3}\end{gathered}

The negative root is not a length. Here $V''=-12\pi h<0$ , and the derivative changes from positive to negative across the allowed root, giving the greatest volume.

Answer

V_{\max}=\frac{500\pi}{3\sqrt3}\text{ cm}^3

If asked for the cylinder’s full height, report $2h=10/\sqrt3$ , not $h$ .

Match corresponding sides in similar triangles

Now a cylinder stands on the base of a cone of height 24 cm and base radius 12 cm. Its top rim touches the cone. Its height is $h$ and radius is $r$ . The small triangle above it and the full half-cone have the same angles, so corresponding side ratios are equal.

Match the small top triangle to the full half-cone: r/12 = (24 − h)/24.

\begin{gathered}\frac{r}{12}=\frac{24-h}{24}\\h=24-2r\end{gathered}

The small triangle’s height is the space above the cylinder, $24-h$ , not $h$ . Consequently $V=\pi r^2(24-2r)$ , with $0<r<12$ .

Your turnPractice

Find the greatest volume of the cylinder in this cone. Give the radius and height, and justify the maximum.

Show worked answer

\begin{aligned}V'&=48\pi r-6\pi r^2\\&=6\pi r(8-r)=0\end{aligned}

$r=0$ is excluded; $r=8$ gives $h=8$ . The derivative changes from positive to negative, so the greatest volume is $512\pi\text{ cm}^3$ . Also $V''(8)=-48\pi<0$ .

Choose the relationship yourself

Your turnPractice

A rectangle has its lower corners on the $x$ -axis and its upper corners at $(-p,9-p^2)$ and $(p,9-p^2)$ , with $p>0$ . Find its greatest area.

Show worked answer

The width is the coordinate difference $2p$ ; the height is $9-p^2$ . Positivity gives $0<p<3$ . Thus $A=18p-2p^3$ , $A'=18-6p^2=0$ , so $p=\sqrt3$ . The derivative changes from positive to negative here: maximum area $12\sqrt3$ square units.

A cross-section can supply the constraint

A cylinder fits inside a sphere of radius 5 cm, with both circular rims touching the sphere. Let its radius be $r$ cm and its full height be $2h$ cm. Cut vertically through the common centre.

The full cylinder height is 2h. The right triangle uses only the upper half, h.

\begin{gathered}V=\pi r^2(2h)\\=50\pi h-2\pi h^3\\0<h<5\\V'=50\pi-6\pi h^2=0\\h=\frac5{\sqrt3}\end{gathered}

The negative root is not a length. Here $V''=-12\pi h<0$ , and the derivative changes from positive to negative across the allowed root, giving the greatest volume.

Answer

V_{\max}=\frac{500\pi}{3\sqrt3}\text{ cm}^3

If asked for the cylinder’s full height, report $2h=10/\sqrt3$ , not $h$ .

Match corresponding sides in similar triangles

Match the small top triangle to the full half-cone: r/12 = (24 − h)/24.

\begin{gathered}\frac{r}{12}=\frac{24-h}{24}\\h=24-2r\end{gathered}

The small triangle’s height is the space above the cylinder, $24-h$ , not $h$ . Consequently $V=\pi r^2(24-2r)$ , with $0<r<12$ .

Your turnPractice

Find the greatest volume of the cylinder in this cone. Give the radius and height, and justify the maximum.

Show worked answer

\begin{aligned}V'&=48\pi r-6\pi r^2\\&=6\pi r(8-r)=0\end{aligned}

$r=0$ is excluded; $r=8$ gives $h=8$ . The derivative changes from positive to negative, so the greatest volume is $512\pi\text{ cm}^3$ . Also $V''(8)=-48\pi<0$ .

Choose the relationship yourself

Your turnPractice

A rectangle has its lower corners on the $x$ -axis and its upper corners at $(-p,9-p^2)$ and $(p,9-p^2)$ , with $p>0$ . Find its greatest area.

Show worked answer