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Make a 5cm deep uncovered rectangular iron box with a 40cm long and 20cm wide rectangular iron sheet. What is the maximum volume of this iron box
Because the depth doesn't change
The area of the beginning and the bottom should be the largest
The bottom is square
Because the width is 20cm
It is known that the four sides of a square are equal
So the length is also 20cm
Then the bottom area is 20 × 20 = 400cm & sup2;
Known volume = bottom area × height
therefore
The volume is 400 × 5 = 2000 CM & sup3;
A 40 cm long and 20 cm wide rectangular sheet iron is welded with a 5 cm deep uncovered cuboid
Let the length and width of a cuboid be x and y, respectively
xy+2*5x+2*5y=40*20
*2+40*20
A rectangular sheet of iron, 20 cm long and 16 cm wide, is welded into an open iron box by subtracting four cm long squares from its four corners? At least how much metal is needed to weld the box?
Solution: (20-4-4) × (16-4-4) × 4 = 12 × 8 × 4 = 384 (cubic centimeter); 20 × 16-4 × 4 × 4 = 320-64 = 256 (square centimeter); answer: the volume of the iron box is 384 cubic centimeter, making such a box needs at least 256 square centimeter iron sheet
The sum of a polynomial and the square of m plus m is the square of M minus 2m
－3m
As shown in the figure, the two high AD and BM of △ ABC intersect at e, and EC, ∠ AEB = 105 ° and ∠ ead = 45 ° are proved as follows: (1) AB = 2am, (2) BC = AC, (3) ab-be = CE, (4) am-cm = CE
As shown in the figure, the two high AD and BM of △ ABC intersect at e, connecting EC, ∠ AEB = 105 ° and ∠ ead = 45 ° to prove: (1) AB = 2am, (2) BC = AC; (3) ab-be = CE; (4) am-cm = CE
Correction: (1) AB = 2am, that is, ABM = 30 & # 186;, should be bad = 45 & # 186; (2) BAM = 60 & # 186;, abd = 45 & # 186;, BC ≠ AC, should be be be = AC
Find a tangent line of the circle x ^ 2 + y ^ 2 = 1, make the area between the tangent line and the parabola y = x ^ 2-2 take the minimum value, and find the minimum value
Let the tangent equation be y = KX + B, where | B | / √ (K & # 178; + 1) = 1, that is, the distance between the tangent and the circle point is equal to 1 (circle radius);
The circle is contained in the parabola, and both of them are symmetric about the y-axis. Therefore, when the area of the interval surrounded by the tangent and parabola is the smallest, b 0 is calculated, B & # 178; = K & # 178; + 1;
Substituting the linear equation into the parabolic equation: KX + B = x & # 178; - 2,
The equation X1 + x2 = k, X1 * x2 = - B-2, x2-x1 = √ [K & # 178; + 4 (B + 2)];
Then s = ∫ {x = x1 → x2} [(KX + b) - (X & # 178; - 2)] DX = (KX & # 178 / 2) - X & # 179 / 3 + (B + 2) x | {x1, X2}
=k(x2²-x1²)/2-(x2³-x1³)/3+(b+2)(x2-x1)
=k(x2-x1)(x2+x1)/2-(x2-x1)[(x1+x2)²-x1*x2]/3+(b+2)(x2-x1)
=k*√[k²+4(b+2)]*k/2-√[k²+4(b+2)]*[k²+(b+2)]/3+(b+2)*√[k²+4(b+2)]；
=√[k²+4(b+2)]*[(k²/2)-(k²+b+2)/3+(b+2)]
=√(b²+4b+7)(b²+4b+7)/6
=[(b²+4b+7)^1.5]/6…… u=√(b²+4b+7)≥√3；
When u = √ 3, s = √ 3 / 2 is the smallest, corresponding to B = - 2, k = ± √ 3;
The area of rectangle ABCD is 80 square centimeters. E and F are the midpoint of AB and ad respectively. Find the area of triangle efc
Let the length of the rectangle be a and the width be B, so AB = 80 square centimeters,
Area of △ EFC = area of rectangle ABCD - △ AEF - △ BCE - △ CDF = AB-1 / 2 * 1 / 2A * 1 / 2b-1 / 2 * 1 / 2A * B-1 / 2 * a * 1 / 2B = 3 / 8ab = 30 square centimeter
It is known that propositions P: X1 and X2 are two roots of the equation x ^ 2-mx-2 = 0
Inequality a ^ 2-5a-3 > = [(x1-x2) absolute value] for any real number m belongs to [- 1,1] holds, proposition q: inequality ax ^ 2 + 2x-1 > 0 has solution, if proposition p is true proposition, q is false proposition, find the value range of A
Why | x1-x2 | = √ (m ^ 2 + 8)? X1 and X2 are the two real roots of the equation x ^ 2-mx-2 = 0,
|x1-x2|=√(m^2+8)
If P is a true proposition, then a ^ 2-5a-3 ≥ √ (m ^ 2 + 8)
Then: (a-5 / 2) ^ 2 ≥ 37 / 4 + √ (m ^ 2 + 8)
If any real number m belongs to [- 1,1], then m ^ 2 ≤ 1
Then: (a-5 / 2) ^ 2 ≥ 37 / 4 + √ (1 + 8)
That is: (a-5 / 2) ^ 2 ≥ 49 / 4
The solution of inequality is a ≤ - 1 and a ≥ 6
Q is a false proposition
Then the opening of the conic ax ^ 2 + 2x-1 is downward, and the maximum value is ≤ 0
The formula of conic is a (x + 1 / a) ^ 2-1-1 / A
When x = - 1 / A, the conic has the maximum value, that is - 1-1 / a ≤ 0
The solution is a ≤ - 1
According to the result of proposition p, the value range of a is a ≤ - 1
Why | x1-x2 | = √ (m ^ 2 + 8)?
Weida theorem
x1+x2=m
x1x2=-2
|x1-x2|=√[(x1+x2)^2-4x1x2]=√[m^2-4*(-2)]=√(m^2+8)
Let o, a, B and C be four points on the plane, and the vector OA = a
Let o, a, B, C be four points on the plane, vector OA = a, vector ob = B, vector OC = C, a + B + C = 0, ab = BC = CA = - 1, then | a | + | B | + | C | =?
help
There is a process to find the value of this polynomial
When Xiao Hua calculates (X & sup2; - 2xy-1) + m, he mistakenly regards "+ m" as "- M" and gets the answer 2x & sup2; - XY + 1. Can you help him find the answer to the original question?
x^2-2xy-1-m=2x^2-xy+1
-->m=x^2-2xy-1-2x^2+xy-1=-x^2-xy-2
So x ^ 2-2xy-1 + M = x ^ 2-2xy-1-x ^ 2-xy-2 = - 3xy-3
Original answer: - 3xy-3
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