Optimisation: maxima and minima in context

a) The sides perpendicular to the wall measure $x$ each; the side parallel to it measures $80-2x$. The area is $A(x)=x(80-2x)=80x-2x^2$, with $0<x<40$.

b) $A^{\prime }(x)=80-4x=0\Rightarrow x=20$. Since $A^{\prime \prime }(x)=-4<0$, it is a maximum.

c) The other side measures $80-40=40$ m, so $A(20)=20\cdot 40=800$ m${}^2$.

a) Volume: $\pi r^{2}h=250\pi$, so $h=\frac{250}{r^2}$. The total area is that of the two lids plus that of the side: $$A=2\pi r^2+2\pi rh=2\pi r^2+2\pi r\cdot \frac{250}{r^2}=2\pi r^2+\frac{500\pi }{r}.$$

b) $A^{\prime }(r)=4\pi r-\frac{500\pi }{r^2}=0\Rightarrow 4\pi r^3=500\pi \Rightarrow r^3=125\Rightarrow r=5$. To justify the minimum: $A^{\prime \prime }(r)=4\pi +\frac{1000\pi }{r^3}>0$ for all $r>0$, so $r=5$ is a minimum. The height is $h=\frac{250}{25}=10$ cm.

c) $A(5)=2\pi \cdot 25+\frac{500\pi }{5}=50\pi +100\pi =150\pi$ cm${}^2$.

a) The perimeter is: two vertical sides ($2h$), the base ($2r$) and the arc ($\pi r$). So $2h+2r+\pi r=6$, and $$h=\frac{6-(2+\pi )r}{2}.$$

b) The area is that of the rectangle ($2r\cdot h$) plus that of the semicircle ($\frac{1}{2}\pi r^2$): $$A=2rh+\frac{\pi r^2}{2}=r(6-(2+\pi )r)+\frac{\pi r^2}{2}=6r-(2+\pi )r^2+\frac{\pi r^2}{2}=6r-\left(2+\frac{\pi }{2}\right)r^2.$$

c) Domain: $h>0$ requires $0<r<\frac{6}{2+\pi }$. $A^{\prime }(r)=6-(4+\pi )r=0\Rightarrow r=\frac{6}{4+\pi }\approx 0.84$ m, which lies in the domain. $A^{\prime \prime }(r)=-(4+\pi )<0$, so it is a maximum. Maximum area: $A=6r-\frac{4+\pi }{2}{r}^2=6r-3r=3r=\frac{18}{4+\pi }\approx 2.52$ m${}^2$ (using $(4+\pi )r=6$).

d) With $r=\frac{6}{4+\pi }$ we get $h=\frac{6-(2+\pi )r}{2}=\frac{1}{2}\cdot \frac{6(4+\pi )-6(2+\pi )}{4+\pi }=\frac{6}{4+\pi }$. Therefore $h=r$ and the ratio is $1$.

In the exam: the objective function, the derivative, the critical value and the justification of the maximum are marked separately. Also write the domain: without it, a critical value outside the allowed range could go unnoticed.