(a) A square shaped thin material with area 196 sq. units to make into an open box by cutting small equal squares from the four corners and folding the sides upward. Prove that the length of the side of a removed square is $\dfrac{7}{3}$ when the volume of the box is maximum. **OR** (b) If F is the constant force generated by the motor of an automobile of mass M, its velocity V is given by $M\dfrac{dV}{dt}=F-kV$, where k is a constant. Prove that $V=\dfrac{F}{k}\left(1-e^{\frac{-kt}{M}}\right)$ when $t=0$ and $V=0$.

> [!TLDR] > (a) sets up the open-box volume as a function of the cut-square side $x$ and uses the first/second derivative test to prove the maximizing value is $x=7/3$; (b) solves the linear first-order ODE $M\,dV/dt=F-kV$ by separation of variables with the initial condition $V(0)=0$. > > **(a) Maximum-volume open box** > 1. Square sheet area $=196$ sq. units $\Rightarrow$ side $=\sqrt{196}=14$ units. > 2. Cutting a square of side $x$ from each corner and folding up: base side $=14-2x$, height $=x$, so $V(x)=x(14-2x)^2$, for $0 3. Expand: $V(x)=x(196-56x+4x^2)=196x-56x^2+4x^3$. > 4. $V'(x)=196-112x+12x^2$. Set $V'(x)=0$: $12x^2-112x+196=0\Rightarrow3x^2-28x+49=0$ (dividing by 4). > 5. $x=\dfrac{28\pm\sqrt{28^2-4(3)(49)}}{2(3)}=\dfrac{28\pm\sqrt{784-588}}{6}=\dfrac{28\pm\sqrt{196}}{6}=\dfrac{28\pm14}{6}$, giving $x=7$ or $x=\dfrac73$. > 6. $x=7$ makes the base side $14-2(7)=0$ (degenerate, no box), so it is rejected. The valid critical point is $x=\dfrac73$. > 7. $V''(x)=-112+24x$. At $x=\dfrac73$: $V''=-112+24\left(\dfrac73\right)=-112+56=-56 8. Hence the volume is maximum when the side of the removed square is $x=\dfrac73$, as required to prove. > > **(b) Solving $M\dfrac{dV}{dt}=F-kV$** > 1. Separate variables: $\dfrac{dV}{F-kV}=\dfrac{dt}{M}$. > 2. Integrate both sides: $-\dfrac1k\ln|F-kV|=\dfrac{t}{M}+C$. …

(a) A square shaped thin material with area 196 sq. units to make into an open box by cutting small equal squa

Question 123 of 148

\dfrac{7}{3}

(b) If F is the constant force generated by the motor of an automobile of mass M, its velocity V is given by

M\dfrac{dV}{dt}=F-kV

, where k is a constant. Prove that

V=\dfrac{F}{k}\left(1-e^{\frac{-kt}{M}}\right)

when

t=0

and

V=0

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Concept understanding — Maxima and Minima

Absolute (global) extrema. For $f$ defined on a domain $D$ , $f(x_0)$ is the absolute maximum of $f$ on $D$ if $f(x_0)\ge f(x)$ for every $x\in D$ ; the absolute minimum is defined symmetrically with $\le$ .

Extreme Value Theorem. If $f$ is continuous on a closed interval $[a,b]$ , then $f$ attains both an absolute maximum and an absolute minimum somewhere on $[a,b]$ — and the extremum can only occur either at an interior critical number or at one of the two endpoints.

Procedure for absolute extrema on $[a,b]$ (Exercise 7.6 Q1's method):

Find every critical number of $f$ in the open interval $(a,b)$ .
Evaluate $f$ at each critical number and at both endpoints $a,b$ .
The largest of these values is the absolute maximum; the smallest is the absolute minimum.

Relative (local) extrema. $f$ has a relative (local) maximum at $x_0$ if $f(x_0)$ is the largest value of $f$ on some open interval around $x_0$ (relative minimum: smallest, on some open interval). A function may have several local extrema, and a local extremum need not be the absolute one.

Fermat's Theorem. If $f$ has a relative extremum at $x=c$ , then $c$ must be a critical number of $f$ (so the search for local extrema always starts by solving $f'(x)=0$ together with any points where $f'$ fails to exist) — though not every critical number is automatically an extremum (e.g. $y=x^3$ at $x=0$ ).

First Derivative Test. At a critical point $c$ where $f$ is continuous, examine the sign of $f'(x)$ moving left to right across $c$ :

negative $\to$ positive: local minimum at $c$ ;
positive $\to$ negative: local maximum at $c$ ;
no sign change (same sign on both sides): $c$ is neither a local max nor a local min.

Second Derivative Test (an alternative, often quicker, at a stationary point). If $f'(c)=0$ and $f''(c)$ exists:

$f''(c)<0$ $\Rightarrow$ local maximum at $c$ ;
$f''(c)>0$ $\Rightarrow$ local minimum at $c$ ; …

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(a) sets up the open-box volume as a function of the cut-square side $x$ and uses the first/second derivative test to prove the maximizing value is $x=7/3$ ; (b) solves the linear first-order ODE $M\,dV/dt=F-kV$ by separation of variables with the initial condition $V(0)=0$ .

(a) Maximum-volume open box

Square sheet area $=196$ sq. units $\Rightarrow$ side $=\sqrt{196}=14$ units.
Cutting a square of side $x$ from each corner and folding up: base side $=14-2x$ , height $=x$ , so $V(x)=x(14-2x)^2$ , for $0<x<7$ .
Expand: $V(x)=x(196-56x+4x^2)=196x-56x^2+4x^3$ .
$V'(x)=196-112x+12x^2$ . Set $V'(x)=0$ : $12x^2-112x+196=0\Rightarrow3x^2-28x+49=0$ (dividing by 4).
$x=\dfrac{28\pm\sqrt{28^2-4(3)(49)}}{2(3)}=\dfrac{28\pm\sqrt{784-588}}{6}=\dfrac{28\pm\sqrt{196}}{6}=\dfrac{28\pm14}{6}$ , giving $x=7$ or $x=\dfrac73$ .
$x=7$ makes the base side $14-2(7)=0$ (degenerate, no box), so it is rejected. The valid critical point is $x=\dfrac73$ .
$V''(x)=-112+24x$ . At $x=\dfrac73$ : $V''=-112+24\left(\dfrac73\right)=-112+56=-56<0$ , confirming a maximum.
Hence the volume is maximum when the side of the removed square is $x=\dfrac73$ , as required to prove.

(b) Solving $M\dfrac{dV}{dt}=F-kV$

Separate variables: $\dfrac{dV}{F-kV}=\dfrac{dt}{M}$ .
Integrate both sides: $-\dfrac1k\ln|F-kV|=\dfrac{t}{M}+C$ . …

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Q.(a) A square shaped thin material with area 196 sq. units to make into an open box by cutting small equal squares from the four corners and folding the sides upward. Prove that the length of the side of a removed square is 73\dfrac{7}{3}37​ when the volume of the box is maximum. OR

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Q.(a) A square shaped thin material with area 196 sq. units to make into an open box by cutting small equal squares from the four corners and folding the sides upward. Prove that the length of the side of a removed square is $\dfrac{7}{3}$ when the volume of the box is maximum. OR

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