The stationary point equations are $x^2+2y-21/4=0$ and $y^2+2x-21/4=0$. Subtracting them factors as $(x-y)(x+y-2)=0$, so all solutions lie either on $x=y$ or on $x+y=2$. The first case gives $x=y=3/2$ or $x=y=-7/2$; the second gives $(x,y)=(-1/2,5/2)$ or $(5/2,-1/2)$. This also verifies the two stated points and exhausts the possible solutions.

The Hessian matrix is $H=\left(\begin{array}{cc}2x& 2\\ 2& 2y\end{array}\right)$, with determinant $4(xy-1)$. The second-derivative test gives:

These extrema are local. Along $x=y=t$ the cubic term makes the function unbounded above and below, so it has no global maximum or minimum on the whole plane.

The base costs $ab$, the glass front costs $2ac$, the wooden rear costs $ac$, and the other two walls cost $2bc$. Thus minimize $C=ab+3ac+2bc$ subject to $abc=V>0$. With the Lagrange multiplier convention $\nabla C=\lambda \nabla (abc)$, the equations areMultiply these by $a,b,c$, respectively. Comparing their equal right sides gives $ab=3ac=2bc$. Since all dimensions are positive, $a=2c$ and $b=3c$. The volume condition then yieldsThis stationary shape is the unique global minimum, not merely a candidate: the arithmetic-geometric mean inequality applied to the three cost terms giveswith equality exactly at those ratios. Therefore the least cost, in pounds in the specified units, isThis is a weighted open-box minimization, with each pair of opposite walls combined into its appropriate positive face weight.

For equation (1), the characteristic equation is $r^2+3r+2=(r+1)(r+2)=0$, so $y=A/x+B/x^2$. The initial conditions give $A+B=0$ and $-A-2B=1$, whence $A=1$, $B=-1$. ThusFor equation (2), the transformed solution is $Y=C\cos z+D\sin z$. At $x=1$ one has $z=0$ and $y^{\prime }(1)=Y^{\prime }(0)$. The initial conditions therefore give $C=0$, $D=1$, yieldingThe domain $x>0$ is the regular interval containing the initial point and the one covered by $x=e^z$.

For $x=e^z>0$, let $Y(z)=y(e^z)$. The chain rule givesMultiplying each original Cauchy-Euler equation by $x^2$ therefore givesandThe exponential change of variable has made both equations have constant coefficients.

The function $x^{-3}$ is strictly decreasing for positive $x$. On $a<x<a+1$, this gives $x^{-3}<a^{-3}<(x-1{)}^{-3}$. Integrate across that unit interval to obtainApplying this to every integer $n\ge a$ and summing proves the integral test tail boundsBoth remain strict: the sum of the positive differences between rectangles and integrals is already bounded below by its positive first difference. Convergence of the integrals and the p-series makes the passage to infinity valid.

Choose a common origin at vertex $C$, set $\mathbf{a}=\overrightarrow{CB}$ and $\mathbf{b}=\overrightarrow{CA}$, and orient the remaining edge as $\mathbf{c}=\overrightarrow{AB}$. ThenThis convention makes the interior angle between $\mathbf{a}$ and $\mathbf{b}$ equal to $C$, and their lengths equal to the opposite side lengths $a$ and $b$.

Take the dot product of the difference with itself. Since $\mathbf{a}\cdot \mathbf{b}=ab\cos C$, the law of cosines follows:The magnitude of the cross product of two edge vectors is twice the triangle's area $K$. Applying this at each vertex gives $2K=ab\sin C=bc\sin A=ca\sin B$. Divide by the nonzero product $abc$ to obtain the law of sines:

The hyperbolic cosine and hyperbolic sine are defined by $\boxed{\cosh x=(e^x+e^{-x})/2,\sinh x=(e^x-e^{-x})/2}$. Their quotient is $\tanh x=\sinh x/\cosh x$.