## Task B-2.7.

From a rectangular plate with side lengths of $12 \mathrm{~cm}$ and $7 \mathrm{~cm}$, one corner has been cut off in the shape of a right-angled triangle with legs of $4 \mathrm{~cm}$ and $3 \mathrm{~cm}$, as shown in the figure. A new rectangular plate is to be cut from the remaining part. What is the maximum possible area of the new rectangular plate? Determine its dimensions (side lengths).

![](https://cdn.mathpix.com/cropped/2024_05_30_63d9d2269da4cabd8f72g-14.jpg?height=325&width=512&top_left_y=263&top_left_x=1206)
## Solution.

Let's denote $x$ and $y$ as shown on the left image. The lengths of the sides of the new rectangle are $12-x$ and $7-y$, so its area is $P=(12-x)(7-y)$.
![](https://cdn.mathpix.com/cropped/2024_05_30_63d9d2269da4cabd8f72g-14.jpg?height=430&width=1214&top_left_y=884&top_left_x=320)

Due to the similarity of the triangles (on the right image), we have $\frac{3}{4}=\frac{y}{4-x}$, which implies $y=3-\frac{3}{4} x . \quad 2$ points

Then,

$$
P=P(x)=(12-x)\left(4+\frac{3}{4} x\right)=-\frac{3}{4} x^{2}+5 x+48
$$

The maximum value of the quadratic function $P$ is achieved at

$$
x=-\frac{5}{2 \cdot\left(-\frac{3}{4}\right)}=\frac{10}{3} \quad \Longrightarrow \quad y=3-\frac{3}{4} \cdot \frac{10}{3}=\frac{1}{2}
$$

The maximum value of the quadratic function $P$ is

$$
P_{0}=\frac{4 \cdot\left(-\frac{3}{4}\right) \cdot 48-5^{2}}{4 \cdot\left(-\frac{3}{4}\right)}=\frac{169}{3}
$$

(that is, $\left.P\left(\frac{10}{3}\right)=\frac{169}{3}\right)$.

The dimensions of the new plate are $12-x=12-\frac{10}{3}=\frac{26}{3} \mathrm{~cm}, 7-y=\frac{13}{2} \mathrm{~cm}$.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

3rd grade - high school - B variant
January 4, 2018.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.