= Solution
Capital and labour are nonnegative inputs. On the budget line, $0\leq K\leq b$ and $L=(b-K)/w$. Both endpoints give zero output, whereas every interior point has positive output, so the maximum is interior. The <Lagrangian> with the sign convention of part (i) is
$$
\mathcal L(K,L,\lambda)=K^\alpha L^\beta+\lambda(b-K-wL).
$$
Its stationarity equations are $\alpha K^{\alpha-1}L^\beta=\lambda$ and $\beta K^\alpha L^{\beta-1}=w\lambda$. Dividing by the positive output gives $\alpha/K=\lambda/f$ and $\beta/L=w\lambda/f$, hence $wL=(\beta/\alpha)K$. Together with the budget constraint this yields
$$
\boxed{K_* =\frac{\alpha b}{\alpha+\beta},\qquad L_* =\frac{\beta b}{w(\alpha+\beta)}.}
$$
To verify the unique global maximum, maximize the logarithm of the <Cobb–Douglas production function> along the budget line:
$$
\log f=\alpha\log K+\beta\log(b-K)-\beta\log w,\qquad\frac{d^2}{dK^2}\log f=-\frac\alpha{K^2}-\frac\beta{(b-K)^2}<0.
$$
Its unique stationary point is therefore the unique maximum, including comparison with the zero-output boundary points.
Writing $r=\alpha+\beta$ and $C=(\alpha/r)^\alpha(\beta/(wr))^\beta$, the optimal output is $\phi(b)=Cb^r$. The <Lagrange multiplier> obtained from the first stationarity equation is $\lambda=\alpha\phi/K_*=r\phi/b$. Direct differentiation verifies the <envelope theorem> identity:
$$
\boxed{\phi(b)=\left(\frac\alpha{\alpha+\beta}\right)^\alpha\left(\frac\beta{w(\alpha+\beta)}\right)^\beta b^{\alpha+\beta},\qquad\phi'(b)=\lambda(b)=\frac{(\alpha+\beta)\phi(b)}b.}
$$
For clarity, the global dual-supremum premise of part (i) need not hold for every exponent pair allowed here. If $r>1$, along $(K,L)=(tK_*,tL_*)$ the unrestricted <Lagrangian> is $t^r\phi(b)+\lambda b(1-t)$, which tends to infinity for every finite $\lambda$. Thus the verification above is by constrained stationarity and direct differentiation, and remains valid even when that stronger global <Lagrangian duality> representation fails.
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