Q.Find two positive numbers $x$ and $y$ such that $x + y = 60$ and $xy^3$ is maximum.

Concept and Intuition

For products of powers like $x^\alpha(a-x)^\beta$, logarithmic differentiation converts the product into a sum, making the derivative condition for a maximum a simple linear equation in $x$. The general result is that the maximum occurs at $x=\dfrac{a\alpha}{\alpha+\beta}$ — a weighted average of the endpoints.

Step-by-Step Solution

1. Let $f(x)=x^\alpha(24-x)^\beta$. Take logs: $\log f = \alpha\log x + \beta\log(24-x)$.
2. Differentiate: $\dfrac{f'}{f} = \dfrac{\alpha}{x} - \dfrac{\beta}{24-x}$.
3. At a maximum, $f'=0$ (and $f\ne 0$ in the interior), so $\dfrac{\alpha}{x}=\dfrac{\beta}{24-x}$. …

Concept and Intuition

$P$ and $Q$ are opposite ends of a diameter of the circle centred at the origin, so if $P=(a\cos\theta,a\sin\theta)$ then $Q=(-a\cos\theta,-a\sin\theta)$. The perpendicular distance of a point $(x_0,y_0)$ from the line $x+y-1=0$ is $\dfrac{|x_0+y_0-1|}{\sqrt2}$. Writing $u=a(\cos\theta+\sin\theta)$, we get $s=\dfrac{|u-1|}{\sqrt2}$ (from $P$) and $t=\dfrac{|u+1|}{\sqrt2}$ (from $Q$), and $u$ sweeps the interval $[-a\sqrt2,\,a\sqrt2]$ as $\theta$ varies.

Step-by-Step Solution

1. $st=\dfrac{|u-1||u+1|}{2}=\dfrac{|u^2-1|}{2}$.
2. As a function of $u$, $|u^2-1|$ equals $1-u^2$ for $|u|\le 1$ (decreasing away from $u=0$) and $u^2-1$ for $|u|\ge1$ (increasing as $|u|$ grows). So on the allowed range $u\in[-a\sqrt2,a\sqrt2]$, the candidates for the maximum are the interior point $u=0$ and the two boundary values $u=\pm a\sqrt2$. …

Concept and Intuition

With $x+y$ fixed, maximizing $x^3y$ is a standard single-variable calculus optimization after substitution. The twist is recognizing that $k/(\alpha^2\beta^2)$ simplifies algebraically to $\alpha/\beta$ once you know $k=\alpha^3\beta$, without needing the numeric value of $k$ at all.

Step-by-Step Solution

1. Let $f(x)=x^3(20-x)=20x^3-x^4$ for $0<x<20$.
2. $f'(x)=60x^2-4x^3=4x^2(15-x)$. Setting $f'(x)=0$ (and $x\ne0$) gives $x=15$.
3. Check it's a maximum: $f''(x)=120x-12x^2$; at $x=15$, $f''=120(15)-12(225)=1800-2700=-900<0$, confirming a maximum.
4. So $\alpha=15$, $\beta=20-15=5$, and $k=f(15)=\alpha^3\beta$. …

Concept and Intuition

Instead of Lagrange multipliers, a constraint of the form (sum of two non-negative terms) $=$ constant, with the two terms' product proportional to $(xy)^2$, is a textbook AM–GM optimisation.

Step-by-Step Solution

1. Given $a^2x^4+b^2y^4=c^6$, with $a^2x^4\ge 0,\ b^2y^4\ge 0$.
2. By AM–GM: $a^2x^4+b^2y^4 \ge 2\sqrt{a^2x^4\cdot b^2y^4} = 2ab\,x^2y^2$.
3. So $c^6 \ge 2ab(xy)^2 \Rightarrow (xy)^2 \le \dfrac{c^6}{2ab}$. …

Concept and Intuition

A line with $x$-intercept $a$ and $y$-intercept $b$ forms a right triangle with the axes of area $\tfrac12ab$. If $a+b$ is fixed, then by AM–GM (or simple calculus), $ab$ — and hence the area — is maximized when $a=b$.

Step-by-Step Solution

1. Let intercepts be $a,b$ with $a+b=12$. Area $=\tfrac12 ab$.
2. Maximize $ab$ subject to $a+b=12$: write $b=12-a$, so $ab=a(12-a)=12a-a^2$. Differentiating: $12-2a=0\Rightarrow a=6$, hence $b=6$ too (this also matches AM–GM equality condition).
3. Maximum area $=\tfrac12(6)(6)=18$.
4. The corresponding line with intercepts $(6,6)$ is $\dfrac x6+\dfrac y6=1 \Rightarrow x+y=6$. …

Concept and Intuition

For two non-negative numbers with fixed sum $S$, their product is maximised when they're equal (a direct consequence of AM ≥ GM, with equality at equal terms). Here the "two numbers" are $a^2x^4$ and $b^2y^4$, whose sum is fixed at $c^6$; their product is $a^2b^2\cdot x^4y^4$, which is exactly $a^2b^2$ times the quantity we want to maximise.

Step-by-Step Solution (AM-GM)

1. Let $u=a^2x^4,\ v=b^2y^4$, so $u+v=c^6$ (fixed), and $x^4y^4=\dfrac{uv}{a^2b^2}$.
2. By AM-GM, $uv$ is maximised when $u=v=\dfrac{c^6}2$, giving $uv_{\max}=\left(\dfrac{c^6}2\right)^2=\dfrac{c^{12}}4$.
3. So $x^4y^4\Big|_{\max}=\dfrac{c^{12}/4}{a^2b^2}=\dfrac{c^{12}}{4a^2b^2}$.

Step-by-Step Solution (Lagrange, as a check)

1. Maximise $f=x^4y^4$ subject to $g=a^2x^4+b^2y^4-c^6=0$. $\nabla f=\lambda\nabla g$ gives $4x^3y^4=\lambda\cdot4a^2x^3$ and $4x^4y^3=\lambda\cdot4b^2y^3$, i.e. $y^4=\lambda a^2$ and $x^4=\lambda b^2$. …

Concept and Intuition

With $x+y+z$ fixed and an ordering constraint $x\ge y\ge z\ge\pi/12$, the feasible set of $(x,y,z)$ is a compact 2-dimensional region (a triangle in the plane $x+y+z=\pi/2$). A smooth function on such a region attains its extreme values either at an interior critical point or on the boundary (vertices/edges). Checking shows there's no valid interior critical point here, so we examine the three vertices of the region.

Step-by-Step Solution

1. The three vertices of the region (obtained by making two of the inequality constraints tight together with the sum constraint) are:
   • $x=y=z=\pi/6$ (all equal),
   • $x=y=\tfrac{5\pi}{24},\ z=\pi/12$,
   • $x=\pi/3,\ y=z=\pi/12$.
2. Evaluate $f=\cos x\sin y\cos z$ at each: at $(\pi/6,\pi/6,\pi/6)$, $f=\tfrac{\sqrt3}{2}\cdot\tfrac12\cdot\tfrac{\sqrt3}{2}=\tfrac38$. At $(\tfrac{5\pi}{24},\tfrac{5\pi}{24},\pi/12)$, $f\approx0.467$. At $(\pi/3,\pi/12,\pi/12)$: $\cos60^\circ=\tfrac12$, and $\sin15^\circ\cos15^\circ=\tfrac12\sin30^\circ=\tfrac14$, so $f=\tfrac12\cdot\tfrac14=\tfrac18$. …

Concept and Intuition

With the constraint $xy=6$, we can express $y$ in terms of $x$ and reduce to a single-variable minimization, or recognize that AM-GM directly bounds $2x+3y$ below in terms of the fixed product $xy$.

Step-by-Step Solution

1. From $xy=6$, $y=\dfrac{6}{x}$ (with $x>0$ so that $y>0$ too, both positive since the product is positive).
2. $f(x)=2x+3y=2x+\dfrac{18}{x}$.
3. $f'(x)=2-\dfrac{18}{x^2}$. Setting $f'(x)=0$: $x^2=9\Rightarrow x=3$.
4. $f''(x)=\dfrac{36}{x^3}>0$ at $x=3$, so this is a minimum.
5. $f(3)=2(3)+\dfrac{18}{3}=6+6=12$. …

Concept and Intuition

For a rectangle with sides $x$ and $y$, perimeter $2(x+y)=20\Rightarrow x+y=10$ is fixed, while area $xy$ is to be maximized. By AM–GM, $xy\le\left(\dfrac{x+y}{2}\right)^2$, with equality when $x=y$ — so the maximum-area rectangle for a fixed perimeter is always a square.

Step-by-Step Solution

1. Perimeter constraint: $2(x+y)=20\Rightarrow x+y=10$.
2. Area $A=xy=x(10-x)=10x-x^2$.
3. $\dfrac{dA}{dx}=10-2x=0\Rightarrow x=5$; then $y=10-5=5$ (a square).
4. Maximum area $A=5\times5=25$. …