A box open from top is made from a rectangular sheet of dimension $a \times b$ by cutting squares each of side $x$ from each of the four corners and folding up the flaps. If the volume of the box is maximum, then $x$ is equal to: (1) $\frac { a + b + \sqrt { a ^ { 2 } + b ^ { 2 } - a b } } { 6 }$ (2) $\frac { a + b - \sqrt { a ^ { 2 } + b ^ { 2 } - a b } } { 12 }$ (3) $\frac { a + b - \sqrt { a ^ { 2 } + b ^ { 2 } - a b } } { 6 }$ (4) $\frac { a + b - \sqrt { a ^ { 2 } + b ^ { 2 } + a b } } { 6 }$

Let $x , y > 0$. If $x ^ { 3 } y ^ { 2 } = 2 ^ { 15 }$, then the least value of $3 x + 2 y$ is
(1) 30
(2) 32
(3) 36
(4) 40

Consider a cuboid of sides $2 x , 4 x$ and $5 x$ and a closed hemisphere of radius $r$. If the sum of their surface areas is constant $k$, then the ratio $x : r$, for which the sum of their volumes is maximum, is
(1) $2 : 5$
(2) $19 : 45$
(3) $3 : 8$
(4) $19 : 15$

A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is equal to (if the full question text was truncated, this is the standard formulation of this problem).

Let $S$ be the set of all $a \in N$ such that the area of the triangle formed by the tangent at the point $P(b, c)$, $b, c \in N$, on the parabola $y^2 = 2ax$ and the lines $x = b$, $y = 0$ is 16 unit$^2$, then $\sum_{a \in S} a$ is equal to $\_\_\_\_$.

A wire of length 20 m is to be cut into two pieces. A piece of length $\ell_1$ is bent to make a square of area $A_1$ and the other piece of length $\ell_2$ is made into a circle of area $A_2$. If $2A_1 + 3A_2$ is minimum then $\pi\ell_1 : \ell_2$ is equal to:
(1) 6:1
(2) $3:1$
(3) $1:6$
(4) $4:1$

A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in $\mathrm{cm}^2$) is equal to
(1) 800
(2) 675
(3) 1025
(4) 900

The maximum area of a triangle whose one vertex is at $( 0,0 )$ and the other two vertices lie on the curve $y = - 2 x ^ { 2 } + 54$ at points $( x , y )$ and $( - x , y )$ where $\mathrm { y } > 0$ is :
(1) 88
(2) 122
(3) 92
(4) 108

Let A be the region enclosed by the parabola $y ^ { 2 } = 2 x$ and the line $x = 24$. Then the maximum area of the rectangle inscribed in the region A is $\_\_\_\_$

Consider the region $R = \left\{(x, y) : x \leq y \leq 9 - \frac{11}{3}x^2,\, x \geq 0\right\}$. The area of the largest rectangle of sides parallel to the coordinate axes and inscribed in $R$, is:
(1) $\frac{730}{119}$
(2) $\frac{625}{111}$
(3) $\frac{821}{123}$
(4) $\frac{567}{121}$

Consider all segments PQ of length 2 such that the end points P and Q are on the parabola $y = x ^ { 2 }$. Denote the mid-point of the segment PQ by M. Among all M, we are to find the coordinates of the ones nearest to the $x$-axis.
Let us denote the coordinates of the end points of segment PQ by $\mathrm{ P }\left( p , p ^ { 2 } \right)$ and $\mathrm{ Q }\left( q , q ^ { 2 } \right)$. Then the $y$-coordinate $m$ of M is
$$m = \frac { p ^ { 2 } + q ^ { 2 } } { \mathbf { M } } .$$
Next, since $\mathrm{ PQ } = 2$, then
$$( p - q ) ^ { 2 } + \left( p ^ { 2 } - q ^ { 2 } \right) ^ { 2 } = \mathbf { N }$$
by the Pythagorean theorem.
Now, when we set $t = p q$, we obtain from (1) and (2) the quadratic equation in $m$
$$\mathbf { O } m ^ { 2 } + m - \mathbf { P } t ^ { 2 } - t - \mathbf { Q } = 0 .$$
When we solve this for $m$, noting that $m > 0$, we have
$$m = - \frac { 1 } { \mathbf { R } } + \sqrt { \left( t + \frac { 1 } { \mathbf { S } } \right) ^ { 2 } + \mathbf{T} } .$$
This shows that $m$ is minimized when $t = - \dfrac { 1 } { \mathbf{U} }$. In this case, $p q = - \dfrac { 1 } { \mathbf{U} }$ and $p ^ { 2 } + q ^ { 2 } = \dfrac { \mathbf { V } } { \mathbf { V } }$, and so we have $p + q = \pm \mathbf { W }$.
Thus the coordinates of the M nearest to the $x$-axis are $\left( \pm \dfrac { 1 } { \mathbf { X } } , \dfrac { \mathbf { Y } } { \mathbf { Z } } \right)$.

Consider a rhombus ABCD with sides of length $a$, where $a$ is a constant. Let $r$ be the radius of the circle O inscribed in the rhombus ABCD, and $\mathrm{K}, \mathrm{L}$, $\mathrm{M}, \mathrm{N}$ be the points of tangency of the circle O and the rhombus. Let $S$ denote the area of the part of the rhombus outside circle O.
We are to find the range of the values of $r$, and the maximum value of $S$.
(1) For each of $\mathbf{A}$ $\sim$ $\mathbf{C}$ below, choose the correct answer from among (0) $\sim$ (9).
Let $\angle \mathrm{ABO} = \theta$. We have $\mathrm{OB} = \mathbf{A}$, and hence $\mathrm{OK} = \mathbf{B}$. Hence, since $( \cos \theta - \sin \theta ) ^ { 2 } \geqq 0$, the range of the values taken by $r$ is
$$0 < r \leqq \mathbf { C } .$$
(0) $a$ (1) $\frac { a } { 2 }$ (2) $\frac { a } { 3 }$ (3) $a \sin \theta$ (4) $a \cos \theta$ (5) $a \tan \theta$ (6) $a \sin ^ { 2 } \theta$ (7) $a \cos ^ { 2 } \theta$ (8) $a \sin \theta \cos \theta$ (9) $a \tan ^ { 2 } \theta$
(2) For each of $\mathbf { D } \sim$ $\mathbf{F}$ below, choose the correct answer from among (0) $\sim$ (9).
When the area $S$ is expressed in terms of $r$, we have
$$S = \mathbf { D } .$$
Here, we observe that when $r = \mathbf { E }$, the value of $\mathbf{D}$ is maximized, and this value for $r$ satisfies (1). Thus, at $r = \mathbf { E }$, $S$ takes the maximum value $\mathbf { F }$.
(0) $2 a r - \pi r ^ { 2 }$ (1) $a r - \frac { \pi } { 2 } r ^ { 2 }$ (2) $\frac { a } { 2 } r - \pi r ^ { 2 }$ (3) $\frac { a r - \pi r ^ { 2 } } { 2 }$ (4) $\frac { 2 a } { \pi }$ (5) $\frac { a } { \pi }$ (6) $\frac { a } { 2 \pi }$ (7) $\frac { 4 a ^ { 2 } } { \pi }$ (8) $\frac { a ^ { 2 } } { \pi }$ (9) $\frac { a ^ { 2 } } { 4 \pi }$

Consider an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ in the $xy$-plane. Here, $a$ and $b$ are constants satisfying $a > b > 0$. Answer the following questions.

1. Find the equation of the tangent line at a point $(X, Y)$ on the ellipse in the first quadrant.
2. The tangent line obtained in Question I. 1 intersects the $x$- and $y$-axes. Find the coordinates $(X, Y)$ at the tangent point that minimizes the length of the segment connecting the two intersects and obtain the minimum length of the segment.
3. Consider a region bounded by the segment obtained in Question I. 2 and the $x$- and $y$-axes, and let $C_{1}$ be a cone formed by rotating the region around the $x$-axis. Next, let $C_{2}$ be a cone with the maximum volume while having the same surface area (including a base area) as the cone $C_{1}$. Find $\frac{S_{2}}{S_{1}}$, where $S_{1}$ and $S_{2}$ are the base areas of the cones $C_{1}$ and $C_{2}$, respectively.

A workplace consisting of a corridor, kitchen, and study room has the model shown above as rectangle ABCD, and the perimeter of this rectangle is 72 meters.
For the kitchen in this workplace to have the largest area, what should $x$ be in meters?
A) 1
B) 2
C) 3
D) 4
E) 5

A line $d$ with negative slope passing through the point $(1,2)$ forms a triangular region with the coordinate axes. What is the minimum area of this triangular region in square units?
A) 2
B) 3
C) 4
D) $\frac { 9 } { 2 }$
E) $\frac { 7 } { 2 }$

For $x > 0$; if the point $(a, b)$ on the graph of the curve $y = 6 - x^2$ is closest to the point $(0, 1)$, what is b?
A) $\frac { 3 } { 2 }$
B) $\frac { 5 } { 2 }$
C) $\frac { 7 } { 2 }$
D) $\frac { 9 } { 2 }$
E) $\frac { 11 } { 2 }$

A tour company charges 140 TL per person for a tour it will organize. If the number of registered participants exceeds 80, a refund of 50 kuruş will be made to all participants for each person above 80. The capacity is limited to 200 people.
For example, if 100 people participate in the tour, everyone receives a 10 TL refund and the per-person fee is 130 TL.
Accordingly, how many people should participate in the tour so that the company's revenue from participants is maximum?
A) 160
B) 165
C) 175
D) 180
E) 185

In the rectangular coordinate plane, the graph of the curve $y = e ^ { \left( - x ^ { 2 } \right) }$ is given.
In this plane, a rectangle with one side on the x-axis and two vertices on the curve is drawn with the maximum possible area.
What is the area of this rectangle in square units?
A) $\sqrt { \mathrm { e } }$
B) $\sqrt { 2 e }$
C) $\frac { \sqrt { e } } { 2 }$
D) $\sqrt { \frac { 2 } { \mathrm { e } } }$
E) $2 \sqrt { e }$

In the rectangular coordinate plane, rectangles are drawn such that two vertices lie on the x-axis and the other two vertices lie on the parabola $y = 27 - x ^ { 2 }$, and the rectangles lie between this parabola and the x-axis.
Accordingly, what is the perimeter of the rectangle with the largest area?
A) 40
B) 42
C) 44
D) 46
E) 48

A crystal in the shape of a cube with one edge of length $x$ units has a production cost of 5 TL per unit cube based on volume, and a selling price of 20 TL per unit square based on surface area.
Accordingly, for what value of x in units will the profit from selling this crystal be maximum?\ A) 16\ B) 18\ C) 20\ D) 22\ E) 24