To design a poster, the screen area is required to be 4840 square meters, and the ratio of width to

If the width of the screen is xcm and the height is ycm, then xy=4840

Paper area s=(x+10)(y+16)=xy+16x+10y+1604840+16x+10y+160=16x+10y+50002 (160xy) +5000=6760

When 16x=10y, the minimum value is taken. It is solved by 16x=10y, xy=4840, x=55, y=88

The width and height are a and b respectively, depending on the title.

a= b, picture area s=a*b=4840 m 2 paper area h=(a+

If you have learned the derivative, find the derivative of h, h'=2λb+

Ream'=0, which solves b, at which point b should be the smallest because h'represents the slope of h, when h'When it is 0, the slope of h is 0, that is, h takes the minimum value of the parabola. Then find the value of a.

If you haven't learned the derivative, you can convert h to the form f(x)=(mx-n) 2+p and find the minimum value.

I won't help you with the specifics.

But are you sure you didn't miss a word or two in the unit you gave?

Let the width of the vertical stove code of the sensitive screen be xcm, and the height is ycm, then the remaining xy=4840

Paper area s=(x+10)(y+16)=xy+16x+10y+1604840+16x+10y+160=16x+10y+50002 (160xy) +5000=6760

When 16x=10y, the minimum value is taken. It is solved by 16x=10y, xy=4840, x=55, y=88

If the width of the screen is xcm and the height is ycm, then xy=4840

When the paper area is s=(x+10)(y+16)=xy+16x+10y+160=4840+16x+10y+160=16x+10y+5000 2 (160xy) +5000=676016x=10y, the minimum value is taken. It is solved by 16x=10y, xy=4840, x=55, y=88

Solution: The length and width of the painting are x and y, respectively

Then: x*y=4840

Paper area s=4840+2*5*x+2*8*(y+10)=4840+160+10x+16y=5000+10x+16y

5000 + 2 * [10x*16y] = 5000 + 2 * 880 = 6760

There is a minimum value if and only if 10x=16y.

At this point, x=88, y=55

Solution: The length and width of the painting are x and y, respectively

Then: x*y=4840

Paper area s=4840+2*5*x+2*8*(y+10)=4840+160+10x+16y=5000+10x+16y

5000 + 2 * [10x*16y] = 5000 + 2 * 880 = 6760

There is a minimum value if and only if 10x=16y.

At this point, x=88, y=55

I'm a little confused about this question, so ask some professionals.

According to the meaning of the topic, let the width be b, then a=b· , let the paper area be s, then, s=(a+10)· (b+16), and a·b=b· ·b=b · 4840 (if I understand correctly, the area of the paper is larger than the area of the Qingna propaganda screen), then s counts:

s=4840+16a+10b+160, using the important inequality to obtain s=5000+16a+10b>=5000+2 16a·10b =5000+2 160·4840=5000+880=5880, if and only if the line is old, 16a=10b, take the = sign, then =10 16=5 8

The answer is this, let's see if it's Oh, think about it for a long time, if it's not, it's not good, and hopefully.

Let the height of the picture be xcm, the width is ycm, and there is xy=4840, x 0, y 0 ---2 points) according to the meaning

Then the required paper area s=(x+16)(y+10)=xy+16y+10x+160, that is, s=5000+16y+10x,--4 points).

x 0, Zen fiber y 0, xy=4840

16y+10x≥2160xy

-8 points) if and only if 16y=10x, i.e., x=88, y=55 equal sign is true --9 points) i.e., when the picture height is 88cm, and the width of the cavity is 55cm, the minimum required paper area is 6760cm2

-10 points).

Solution: If the screen height is xcm and the width is xcm, then x24840, and the paper area is s, there is s=(x+16)(x+10)=x2(16+10)x+160, and x=<>

Substituting the above formula, we get s=5000+44<>

When the <> is <>, the minimum value is obtained, and at this time, the high: <>

Width: <>

A: When the screen height is 88 cm and the width is 55 cm, the paper area used can be minimized.

Solution: Let the height of the screen be xcm and the width be xcm, then x2=4840, let the paper area be s, there is s=(x+16)( x+10) = x2+(16 +10)x+160, substitute x= into the above formula, get s=5000+44, when, instant s obtain the minimum value, at this time, height:, width:, answer: when the screen height is 88cm, and the width is 55cm, the paper area used can be minimized.