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Let $F=x+y+z^2$ and $g=x^2+y^2+z^2$. The unit sphere is compact, so the continuous function $F$ has a global maximum and global minimum. Since $\nabla g\ne 0$ on the unit sphere, all such points satisfy the Lagrange multiplier equationsThus $x=y$ and either $z=0$ or $\lambda =1$.

If $z=0$, then $x=y=\pm 1/\sqrt{2}$, giving $F=\pm \sqrt{2}$. If $\lambda =1$, then $x=y=1/2$ and $z=\pm 1/\sqrt{2}$, giving $F=3/2$. These exhaust the candidates, so

Write $x_i\in {\mathbb{R}}^p$ for row $i$ of the data matrix. The k-means clustering objective is the within-cluster sum of squared Euclidean distances:The nonempty $C_k$ form a partition of a set of observation indices. For fixed clusters, part (a) shows that the best $m_k$ is their centroid.

Initialize distinct centres, for example by choosing $K\le n$ distinct observations. Then repeat the following steps:

The empty-cluster repair is possible for $K\le n$ and cannot increase $J$: the transferred observation now has squared distance zero, while all other squared distances stay unchanged before the centre update.

Assignment cannot increase $J$, and any changed assignment strictly decreases it under the tie rule. Updating centres cannot increase $J$ by the centroid sum-of-squares identity. After an update, the centres are determined by the partition of a set. There are only finitely many possible assignments. Unless this is the initial round, any nonterminal round starts with an assignment that strictly reduces $J$; repairing empty clusters and updating cannot undo that decrease. Thus a partition cannot recur, proving finite termination of K-means clustering. The final result need not be a global minimum; different initial centres can lead to different final clusterings.

Expanding and taking partial derivatives gives the gradient and Hessian matrixAt $p_0=(-1,0{)}^T$, these are $\nabla f(p_0)=(-6,2{)}^T$ andThus the Newton step solvesThe second equation gives $s_y=1-2s_x$; substituting into the first gives $22s_x=10$. HenceAt $p_{\ast }=(-1/2,-1/2{)}^T$, however,The eigenvalues of this Hessian matrix are $-1$ in direction $(1,1)$ and $7$ in direction $(1,-1)$. It is invertible but indefinite. Thus the Newton step is zero, andThis illustrates that stationary points are fixed points of Newton's method. Here the fixed point is a saddle point of a scalar function, not a local minimum. Indeed, direct substitution along the two eigenvector directions givesFor sufficiently small nonzero $t$, one direction decreases $f$ and the other increases it. The global minimum is instead zero: the sum of squares vanishes at $(0,0)$ and $(-1,-1)$. These are the only real zeros, since $x=-y^2$ and $y=-x^2$ imply $y(y^3+1)=0$. Thus a zero gradient is an insufficient stopping certificate for minimization problem without classifying the stationary point.

The gradient and constant Hessian matrix areAt ${\mathbf{x}}_0=(1,0{)}^T$, $\nabla F=(10,-8{)}^T$ and $H_F^{-1}\nabla F=(1,-1{)}^T$. Hence one multivariable Newton iteration givesAt this point the gradient vanishes, so every subsequent iterate is $(0,1)$. The one-step result follows because the gradient of a quadratic polynomial is affine and its linearization is exact. Since the leading principal minors of $H_F$ are $6>0$ and $8>0$, it is a positive-definite matrix. Alternatively,directly proves that $(0,1)$ is the unique global minimum, with value $-1$.

Differentiate the ordinary least squares objective in a. Its normal equations for linear least squares areUsing ${\sum }_i{x}_i=0$ and $S_{xx}={\sum }_i{x}_i^2>0$ givesThe Hessian matrix of the objective in $(a,b)$ is $2\mathrm{diag}(n,S_{xx})$, which is positive-definite, so this is its unique global minimum. For the observed estimates, replace $Y_i$ by $y_i$.

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.

Set $g(x,y)=y-(x-2{)}^2$. The gradient of $g$ never vanishes, so the Lagrange multiplier equations $\nabla f=\lambda \nabla g$ apply at a constrained stationary point. They are $4x=-2\lambda (x-2)$, $2y=\lambda$, and $y=(x-2{)}^2$. Eliminating $\lambda ,y$ givesWith $t=x-2$, this becomes $t^3+t+2=(t+1)(t^2-t+2)=0$. The quadratic factor is $(t-1/2{)}^2+7/4>0$, so the only real possibility is $t=-1$. ThereforeThe level curve through this point is the ellipse $2x^2+y^2=3$, meeting the coordinate axes at $(\pm \sqrt{3/2},0)$ and $(0,\pm \sqrt{3})$. Its tangent line at $(1,1)$ has slope $-2x/y=-2$, equal to the derivative $2(x-2)$ of the constraint parabola there. The two curves are tangent.

The parabola touches this ellipse from outside; every other point on the constraint lies on a larger contour. An exact check of this geometric argument iswith equality only at $x=1$. Thus the point is the unique constrained global minimum.

As printed, the function is undefined at the origin. For $r^2=x^2+y^2>0$, its gradient isThe scalar prefactor is positive, and $(x,y)\ne (0,0)$, so this gradient never vanishes. Thus there are no stationary points on the stated domain. Its infimum is three, approached as $r\to 0$, but that value is not attained there.

If the intended definition includes the natural extension $f(0,0)=3$, that extension has a unique strict global minimum at the origin, with value three. Indeed, $e^{-1/r^2}>0$ off zero, and $e^{-1/r^2}/r\to 0$ proves differentiability at zero with zero gradient. More strongly, subtracting three gives the radial Gaussian flat function: every derivative of that difference vanishes at the origin. The vanishing Hessian matrix does not turn this minimum into an inflection point. These are two different domain interpretations; the extension must be stated rather than silently inserted.

Differentiating the hyperbolic cosine with the chain rule gives $g^{\prime }(x)=3{\cosh }^{2}x\sinh x$. For real $x$, $\cosh x>0$, so the only zero of the derivative is $\sinh x=0$, namely $x=0$. ThusSince $\cosh x\ge 1$, this stationary point is the global minimum.

The squared Euclidean distance along the line isIts minimum occurs at $t=0$ exactly when its derivative there vanishes, so $a\cdot (b\times c)=0$. Since $b,c$ are independent, their linear span is the two-dimensional orthogonal complement of $b\times c$. Consequentlyfor unique real $\alpha ,\beta$. This is the closest point on the intersection of two planes; the positive quadratic coefficient $|d{|}^2$ proves that the stationary distance is a strict global minimum.

The flat function $f$ is an even function, has its unique zero and global minimum at the origin, and approaches the horizontal asymptote $1$ as $x\to \pm \infty$. It decreases for negative $x$ and increases for positive $x$, meeting the origin with a horizontal tangent line.

Its derivative is an odd function: $f^{\prime }(x)=2x^{-3}{e}^{-1/x^2}$ off zero, with $f^{\prime }(0)=0$. It is negative to the left and positive to the right, tends to zero both at the origin and at infinity, and has one negative trough and one positive crest. Indeed $f^{\prime \prime }(x)=2e^{-1/x^2}(2-3x^2)/x^6$ away from zero, so the change in slope of $f^{\prime }$ occurs once on each side. The exact crest and trough coordinates are not needed for the sketch.

Use completing the square after taking out the leading coefficient:Thus $a=3$, $b=-2$ and $c=-1$. Since a real square is nonnegative, the global minimum is $-1$, attained at $x=2$.

Minimize the squared distance $F=x^2+y^2$, which has the same minimizers as the distance. The Lagrange multiplier equations for the constraint $G=x^2+xy-1=0$ areThe constraint gradient cannot vanish on this branch, since its second component is $x>0$. Eliminate $\lambda =2y/x$ to obtain $x^2=2xy+y^2$. Put $t=y/x$; then $t^2+2t-1=0$, so $t=-1\pm \sqrt{2}$. But the constraint says $x^2(1+t)=1$, requiring $1+t>0$ and selecting $t=\sqrt{2}-1$. It follows thatTo prove a global minimum on the whole $x>0$ branch, substitute $y=1/x-x$ into $F$:The arithmetic-geometric mean inequality proves this bound, with equality only when $2x^2=x^{-2}$, namely $x=2^{-1/4}$. Thus the stationary point is the unique closest point on the required branch. Numerically it is approximately $(0.840896,0.348311)$ and its distance is $0.910180$ in coordinate units.

For $x>0$, differentiate the two explicit branches:The plus branch is strictly increasing and strictly convex for $x>0$. Its only zero is $x=0$, also its one-sided global minimum; it has no interior extremum or inflection point.

The minus branch rises initially, then has $y_{-}^{\prime }=0$ at $\sqrt{x}=4/5$, changing from positive to negative. Its second derivative changes sign at $\sqrt{x}=8/15$. Its zeros follow from $y_{-}=x^2(1-\sqrt{x})$. ConsequentlyThe origin is a one-sided local minimum of both graph branches, because their domain ends there; it is not an interior critical point. The minus branch has no global minimum since it tends to $-\infty$ as $x\to \infty$. Its maximum and inflection heights, useful for the sketch, are $256/3125$ and $28672/759375$, respectively.

The Taylor series of the logarithm givesThe first nonzero term is therefore $\boxed{x^2}$. It is positive on either side of the origin, so the stationary point is a strict local minimum. In fact $\log (1+x^2)\ge 0$, with equality only at zero, making it the global minimum as well.

Let the fixed positive volume be $V=abc$. For an interior constrained extremum, Lagrange multipliers applied to the energy giveMultiplying these equations by $a,b,c$ shows that $2A/a^2=2A/b^2=2A/c^2=\lambda V$. Since the lengths are positive, $a=b=c=V^{1/3}$.

The arithmetic-geometric mean inequality proves this is the global minimum:with equality exactly for equal sides. Thus the optimal shape is a cube, and $\boxed{E_{\min }=3A{V}^{-2/3}}$.

Put $q=2.4048$ and let the fixed constraint value be $C>0$. For $V=\pi R^{2}H$, the Lagrange multiplier equations giveDividing removes $\lambda$ and yieldsTo establish a minimum, rather than merely a stationary ratio, set $u=q^2/R^2$ and $v={\pi }^2/H^2$. Then $u+v=C$ and $V={\pi }^2{q}^2/(u\sqrt{v})$. The denominator squared is $u^2(C-u)$; its derivative is $u(2C-3u)$, giving a unique maximum at $u=2C/3$, $v=C/3$. The volume tends to infinity at either endpoint. This proves the stated ratio is the unique global minimum, as in the optimal cylinder for a reciprocal-square constraint.

Differentiate the two powers to locate the stationary point:For $x>0$, this is negative before $2$ and positive afterwards, so the global minimum isThe curve crosses the horizontal axis at $x=1$. Its limits are $+\infty$ as $x\to 0^{+}$ and $0$ from below as $x\to \infty$, giving the vertical asymptote $x=0$ and horizontal asymptote $y=0$. These features determine the required sketch.

The Hessian matrix is diagonal:Its eigenvalues are $c$ and $d$. Hence the second-derivative test gives a strict local minimum if $c,d>0$, a strict local maximum if $c,d<0$, and a saddle point of a scalar function if $cd<0$. No degenerate case occurs because both are nonzero.

The local minimum is the unique global minimum, since both separated summands are strictly convex functions and tend to $+\infty$ at either end of the real line. Similarly, when both signs are negative there is a unique global maximum.

The downward drop from the starting rim to angle $\theta$ is $R\cos \theta$. With no dissipative forces, conservation of energy gives $m{v}^2/2=mgR\cos \theta$, so $v=\sqrt{2gR\cos \theta }$. Dividing the line element by this speed yieldsThe endpoint singularities are integrable for the minimizing path. The integrand is independent of $z$; its Euler-Lagrange equation gives a constant $K$, after absorbing the overall prefactor:Solving for $z^{\prime }$ and taking the positive branch required by $z_2>z_1$ givesThe constant is fixed implicitly byThis integral is strictly increasing from zero to infinity as $K$ increases from zero to one, so every finite positive separation selects a unique $K$. The time integrand has second derivative $R^2/[\sqrt{\cos \theta }(R^2+z^{\prime 2}{)}^{3/2}]>0$ with respect to $z^{\prime }$. Its supporting-line inequality, integrated between fixed endpoints with constant first derivative with respect to $z^{\prime }$, proves this stationary path is the unique global minimum within the stated single-valued class.

As $\theta \downarrow -\pi /2$, $z^{\prime }\to 0$. Therefore the skater should initially head directly down the curved ramp, perpendicular to the cylinder axis, and gain axial motion later. The path differs from the constant-pitch shortest route because extra distance is cheaper in time near the faster, lower part of the ramp. This is the brachistochrone on a horizontal half-cylinder.