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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.



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