## Solve Lesson 55 Page 63 SBT Math 10 – Kite>

Topic

Uncle Nam plans to make a rectangular picture frame so that the inner part of the frame is a rectangle with dimensions of 6 cm x 11 cm, the width of the border is $$x$$ cm (Figure 27). The area of ​​the frame border should not exceed $$38c{m^2}$$. What is the width of the largest frame border in cm?

Solution method – See details

Set the width of the frame border to $$x$$(cm) ($$x > 0$$). Representing the area of ​​the frame border and solving the inequalities

Detailed explanation

Set the width of the frame border to $$x$$(cm) ($$x > 0$$).

We have the frame border area $$\left( {11 + 2x} \right)\left( {6 + 2x} \right) – 66 = 4{x^2} + 34x$$ ($$c{) m^2}$$)

According to the problem we have: $$4{x^2} + 34x \le 38 \Leftrightarrow 4{x^2} + 34x – 38 \le 0$$

The quadratic triangle $$4{x^2} + 34x – 38$$ has two solutions $${x_1} = \frac{{ – 19}}{2};{x_2} = 1$$ and has coefficients $$a = 4 > 0$$

Using the sign theorem of quadratic triangles, we see that the set of values ​​of $$x$$ such that the triangle $$4{x^2} + 34x – 38$$ bears the sign “-” is \ (\left[ {\frac{{ – 19}}{2};1} \right]\)

So $$0 < x \le 1$$

So the maximum width of the frame border is 1 cm.